Trigonometry - Neon Classes · 2020. 9. 10. · m Download our app : NEON CLASSES Subscribe our...
Transcript of Trigonometry - Neon Classes · 2020. 9. 10. · m Download our app : NEON CLASSES Subscribe our...
Rose
James Raja
007
Collection
Addition
Partnership Magical
Rex Prime
MatheMagica
Raja
Prime
Partnership
Mantel Maths
Raja
007
Bond
Magical
Rose
James Raja
007
Collection
Addition
Bond Magical
Rex Prime
Partnership
Raja
Prime
Partnership
Mantel Maths
Raja
007
Bond
Magical
Rose
James Raja
007
Collection
Addition
Bond Magical
Rex Prime
Partnership
Raja
Prime
Partnership
Mantel Maths
Raja
007
Bond
Magical
Trigonometry
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1
Contents
S.No. Topic Name Page No.
1. Measurement of Angles 1 – 7
2. Trigometric Functions (T - Ratios) 8 – 21
3. Ratios of standard angles 22 – 33
4. Conversional T – Ratios 34 – 45
5. Basic Identities 46 – 61
6. Sum and difference of angles of T-Ratios 62 - 72
7. Transformation of a Product into a sum or Difference 73 – 86
8. Multiples and sub multiples 87 – 100
9. Special Properties 101 – 115
10. Graphs of Trigonometric Ratios 116 – 125
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1
‘Radians’ and ‘Degrees’ are generally units of
measurement
Why 𝜋 = 180ᵒ:-
Consider a circle of radius r >0, circumference of the
circle c = 2𝜋r where 𝜋 = 22
7
90ᵒ o A B o A o A
B B B
B
180ᵒ 360ᵒ
(a) = 90ᵒ (b) = 180ᵒ (c) = 360ᵒ
Arc AB = 2
4
r arc AB =
2
2
r Arc AB = 2𝜋r
= 2
r = 𝜋r
Here central angle of 360ᵒ cuts off an arc of length 2𝜋r,
which is same as the circumference of the circle. So we
can say
360ᵒ equals 2𝜋r(or 2𝜋 ‘radii’)
Here r is variable and 2𝜋 = constant, so instead of using
the ‘radii’, we use the term radians;
360ᵒ = 2𝜋 radians.
⇒ 1ᵒ = 2
360
radians ⇒ 1ᵒ =
180
radians
♦ Degrees to radians – multiply by 180
x degrees = 180
x
radians
♦ Radians to degree – multiply by 180
x radians = 180
x
radians
or Radians
Degrees 180
=
♦ commonly used angles in degrees
♦ commonly used angles in radians.
♦ Arc length: -
If is the angle subtended at the center of a circle of
radius ‘r’ by an arc of length ‘l’ then
l
r
l
r=
Here l and r are in the same unit and is always in
radians.
Measurement of Angles 1
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2
1. Convert 18ᵒ to radians
Sol. 18ᵒ = 18×180
=
10
or
10
c
2. Convert 9
radians to degrees
Sol. 9
=
9
×
180
= 20ᵒ
3. Convert 11ᵒ15’ to radians
Sol. 1ᵒ = 60 minutes
15’ = 15
60degrees =
1
4degrees
11ᵒ 15’ = 1
114
+
degrees =
45
4degrees
45
4
= 45
4 180 16
= rad
4. Convert 8
to degrees
Sol. 8
=
180
8 =
180
8
= 45
2 = 22.5ᵒ = 22ᵒ
1
2×60’ = 22ᵒ30’
5. Express 45ᵒ20’10’’ in radian measure (𝜋 = 3.14)
Sol. 10’’ = 10
60min. =
10
60 60degrees =
1
360degrees
20’ = 20
60degrees =
1
3degrees
∴ 45ᵒ20’10’’ = 1 1
453 360
+ +
degrees
= 16321
360 degrees
Now
16321 16321
360 360 180
=
rad
= 16321 3.14
360 180 = 0.79 radian
6. Express 1.4 radians in degrees
Sol. 1.4 radian = 1.4×180
degrees
= 1.4×180
22×7 = (80-18)ᵒ = 80ᵒ (.18×60)’
= 80ᵒ (10.8)’ = 80ᵒ10’ (.8×60)’’ = 80ᵒ10’ 48’’
7. Find the length of an arc of a circle of radius 5cm
subtending a central angle measuring 15ᵒ.
Sol. Let l be the length of the arc
Here r = 5cm, = 15ᵒ = 15×180
rad =
12
rad
L = r = 5×5
12 12
= cm
8. Find the length of the arc of a circle of radius 3 inches
subtended by the central angle of 220ᵒ
Sol. let length of arc be l inches
Here radius = 3 inches
= 220ᵒ = 220×11
180 9
= radian
L = r = 3×11 11
9 3
= inches.
9. Find in degrees the angle subtended at the center of the
circle of diameter 30 cm by an arc of length 11cm.
Sol. length of arc = 11cm
Radius = 30
2 = 15 cm
L= r
⇒ = 11
15
l
r= radian
= 11 180
15 degrees
= 11 180
715 22
degrees = 42ᵒ
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3
1. Value of 165ᵒ in radians?
(a) 11
12
(b)
33
35
(c)
5
6
(d)
7
6
2. Degree value of 11
4
radian is
(a) 415ᵒ (b) 390ᵒ (c) 435ᵒ (d) 495ᵒ
3. Convert 49ᵒ to radian
(a) ( )0.73c
(b) ( )0.65c
(c) ( )0.79c
(d) ( )0.85c
4. Which of the following is true?
(a) 415ᵒ = 11
5
(b) 235ᵒ =
43
11
(c) 270 ᵒ = 7
4
(d) 660ᵒ =
11
3
5. Value of 1ᵒ in radian?
(a) 0.1476 radian (b) 0.01746 radian
(c) 0.2836 radian (d) 0.0134
6. Value of 1 radian in degree
=
22
7
(a) 57ᵒ51’22’’ (approx.) (b) 57ᵒ21’16’’ (approx.)
(c) 57ᵒ16’22’’ (approx.) (d) 57ᵒ62,16’’ (approx.)
7. Which of the following is wrong?
(a) 4 2
5 3
+ = 264ᵒ (b)
12 15
+ = 27ᵒ
(c) 7
12 3
− = 45ᵒ (d) None of these
8. Two angles of a triangle are 35ᵒ and 45ᵒ. Third angle is
(a) 5
8
Radian (b)
5
9
radian
(c) 3
7
radian (d)
2
7
radian
9. In a right angled triangle, difference of the two non right
angles is 6
then radian value of both angles is
(a) 2
,
3
(b) ,
4 5
(c) ,
4 6
(d) ,
3 6
10. Two angles of a triangle are 30ᵒ45’15’’ and 29ᵒ14’45’’.
Third angle is
(a) 2
(b)
3
10
(c)
2
3
(d)
12
11. The circular measure of an angle of an isosceles triangle
is5
9
. Find the circular measure of one of the other
angles is -
(a) 5
18
(b)
5
9
(c)
2
9
(d)
4
9
12. Radian value of 63ᵒ14’51’’ is
(a) 2811
8000
radian (b) 3811
8000
radian
(c) 4811
800
radian (d)5811
8000
radian
13. In circular measure, the 30ᵒ30’ is equal to
(a) 51
180
radian (b)
61
180 radian
(c) 61
360 radian (d)
12
14. Radian value of 20ᵒ20’20’’ is
(a) 3601
4320
radian (b)
3661
32400 radian
(c) 2361
3240
radian (d) None of these
15. A central angle in a circle of radius 5cm cuts off an arc of
length 7.5𝜋 cm. What is the measure of angle?
(a) 150ᵒ (b) 270ᵒ (c) 135ᵒ (d) 210ᵒ
16. Find the length of the arc of a circle of radius 10 cm
subtended by central angle of 220ᵒ
Exercise - 1
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(a) 110
9
cm (b)
90
11
cm
(c) 22
9
cm (d) 22𝜋 cm
17. Value of 45ᵒ15’ in radian is
(a) 91
180 radian (b)
91
90 radian
(c) 181
720 radian (d)
181
360 radian
18. The circular measure of the magnitude of the angle of a
regular hexagon
(a) 3
(b)
3
4
(c)
2
3
(d)
5
4
19. if three angles of a quadrilateral are 57ᵒ, 83ᵒ and 115ᵒ
then the fourth angle will be
(a) 3
4
(b)
7
12
(c)
2
3
(d)
4
3
20. If the arc of same length in two circles subtend angles of
60ᵒ and 75ᵒ at their centers. Find the ratio of their radii.
(a) 5 : 4 (b) 4 : 5 (c) 5 : 6 (d) 6 : 5
21. Difference between two angles 5
9
and 54ᵒ35’
(a) 45ᵒ35’ (b) 56ᵒ35’ (c) 46ᵒ25’ (d) 45ᵒ25’
22. Convert 0.6 radians to degrees?
(a) 38ᵒ17’22’’ (b) 35ᵒ16’22’’
(c) 34ᵒ21’49’’ (d) 48ᵒ19’47’’
23. What is the radian measure of 1
8 of a full rotation?
(a) 90ᵒ (b) 45ᵒ (c) 22.5ᵒ (d) 67.5ᵒ
24. Two angles of a triangle are 1
2 radian and
1
3 radian.
The measure of the third angle in degree (taking p =
22/7)
(a) 1
13211
(b) 2
13211
(c) 3
13211
(d) 132°
25. By decreasing 15° of each angle of a triangle, the ratios
of their angles are 2 : 3 : 5. The radian measure of
greatest angle is:
(a) 11
24
(b)
12
(c)
24
(d)
5
24
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Answer Key 1 a 2 d 3 d 4 d 5 b
6 c 7 d 8 b 9 d 10 c
11 c 12 a 13 c 14 b 15 b
16 a 17 c 18 c 19 b 20 a
21 d 22 c 23 b 24 c 25 a
1. (a)
165ᵒ = 165×180
=
11
12
2. (d)
11
4
radian =
11 180
4
= 495ᵒ
3. (d)
49ᵒ = 49×180
radian =
49 22
180 7
=
77
90 = 0.85 rad
4. (d)
415ᵒ = 415×180
=
9
4
radian
235ᵒ = 235×47
180 36
= radian
270ᵒ = 270×180
=
3
2
radian
660ᵒ = 660×11
180 3
= radian
5. (b)
1ᵒ = 180
radian =
22
7 180 radian
= 11
630 radian = 0.01746 radian
6. (c)
1 radian = 180
degree =
1807
22 degrees
= 630
11
= 573
11
= 57ᵒ
'3 60
11
= 57ᵒ16’
''4 60
11= 57ᵒ16’22’’ (approx.)
7. (d)
4 2
5 3
+ = 144ᵒ+120ᵒ = 264ᵒ
12 15
+ = 15ᵒ + 12ᵒ = 27ᵒ
7
12 3
− = 105ᵒ - 60ᵒ = 45ᵒ
8. (b)
Sum of two angles = 35ᵒ+45ᵒ = 80ᵒ
Third angle = 180ᵒ - 80ᵒ
= 100ᵒ = 100ᵒ×180
radian =
5
9
radian
9. (d)
Triangle – right angled
∴ Sum of other two angles = 2
Difference between other two angels = 6
Let other angles be a and b
Here a + b =2
, a – b =
6
∴ a =3
, b =
6
Sum of other two angles should be 2
that is in only
option (d)
10. (c)
Sum of two angles = 30ᵒ45’15’’+29ᵒ14’45’’
= (30+29)ᵒ(45+14)’(15+45)’’ = 59ᵒ59’60’’
= 59ᵒ60’ = 60ᵒ = 3
Third angle = 𝜋 - 2
3 3
=
Solutions
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11. (c)
In isosceles triangle – value of two angles is same.
If both angle are 5
9
then
Sum of two angles = 10
9
> 𝜋 (not possible)
Hence let two equal angels be x radian each
∴ 2x+5
9
= ⇒ x =
2
9
12. (a)
51’’ = 51' 51
60 60 60=
degree =
17
1200 degrees
14’ = 14
60 degrees =
7
30 degrees
63ᵒ14’51’’ =
+ +
7 1763
30 1200 degrees
= 75600 280 17
1200
+ +
= 75897
1200
Now,
75897
1200 180
radian =
2811
8000
radian
13. (c)
30ᵒ 30’ = 30
3060
+
degrees =
130
2
+
degrees
= 61
2 degrees
Now,61
2
=
=61 61
2 180 360 radian
14. (b)
20’’ =
= =
'20 20 1
60 60 60 180
20’ = 20 1
60 3
=
20ᵒ20’20’’ =
+ +
1 120
3 180
=
+ +
=
3600 60 1 3661
180 180
3661
180=
3661
180 180 =
3661
32400 radian
15. (b)
Radius = 5 cm
Length of arc = 7.5𝜋 cm
L = r
⇒ = 7.5
5
l
r
= =
3
2
radian =
3 180
2
= 270ᵒ
16. (a)
Let length of the arc be l cm
l = r
here r = 10 cm, = 220ᵒ = 220×180
=
11
9
L = 10×11
9
=
110
9
17. (c)
45ᵒ15’ = 15
4560
+
=181
4
181 181 181
4 4 180 720
= = radian
18. (c)
Each angle of the regular hexagon = ( )2 180n
n
−
= ( )6 2 180
6
− = 120ᵒ
120 ᵒ = 120×2
180 3
=
19. (b)
Sum of angles of quadrilateral = 360ᵒ
Sum of three angles = 57ᵒ+83ᵒ115ᵒ = 255ᵒ
Fourth angle = 360ᵒ-255ᵒ = 105ᵒ
105ᵒ = 105×7
180 12
= radian
20. (a)
Let r1 and r2 be the radii of given circles,
Length of arcs = same for both = l
For first circle = 60ᵒ = 3
∴ l = r1
r1 =
=3l l
and r2 = 12
5 5
12
l l
=
r1 : r2 = 3 12
:5
l l
= 5 : 4
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21. (d)
5
9
radian =
5 180
9
= 100ᵒ
Now
100ᵒ - 54ᵒ35’ = 45ᵒ25’
22. (c)
0.6 radians = 0.6×180
degrees
= 0.6 180 7 378
22 11
= =
'4 4 60
34 3411 11
=
= 34ᵒ
' ''9 9 60
21 34 21'11 11
=
= 34ᵒ21’49’’
23. (b)
For full rotation = 360ᵒ
Radian measure = 2𝜋
∴ 1
8 of a full rotation =
2
8
=
4
= 45ᵒ
24. (c)
Sum of two angles = 1 1 5
2 3 6+ = radian
Third angle = 5
6
−
radian
= 22 5 132 35 97
7 6 42 42
−− = = rad
radians = 180°
97
42 radians =
180 97
42
= 180 7 97 1455 3
13222 42 11 11
= =
25. (a)
2x + 3x + 5x = 180° – 45° = 135°
10x = 135° x = 135 27
10 2=
Largest angle
= 5x + 15° = 27
52
+ 15° =
135 30 165
2 2
+ =
180° = radian
165 165 11
2 180 2 24
= = radian
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Base (A)
In a right angle triangle
[The side opposite to angle is called perpendicular and
adjacent to angle is called base]
sin = Perpendicular
hypotenuse =
h
p,
L
K
cos = Hypotenuse
Base =
b
h,
A
K
tan = Perpendicular
Base =
p
b,
L
A
cot = 1
tan =
Perpendicular
Base =
b
p,
A
L
sec = 1
cos=
hypotenuse
base =
h
b,
K
A
cosec = 1
sin=
hypotenuse
Perpendicular =
p
h,
K
L
sin = 1
cosec ⇒ sin cosec = 1
cos = 1
sec ⇒ cos sec = 1
tan = 1
cot ⇒ tan cot = 1
sin
cos=
ph
bh
= p
b = tan
= =
cos
sin
bbh
p ph
= cot
Triplets to remember
3 4 5
6 8 10
9 12 15
5 12 13
10 24 26
7 24 25
9 40 41
11 60 61
Trigometric Functions (T-ratios) 2
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1. Find value of sin , cos tan
12
5
Sol. Here perpendicular (L) = 5
Base (A) = 12
Hypotenuse (K) = 13
sin = =5
13
L
K
cos = =12
13
A
K
tan = =5
12
L
A
2. If sin = 11
15, then find tan
Sol.
11
sin =11
15,
sin =L
K
∴ L = 11, K = 15, A = ?
By Pythagoras theorem
K2 = L2+A2 ⇒ A2 = K2 – L2
⇒ A2 = (15)2 – (11)2 ⇒ A2 = 225 – 121
⇒ A2 = 104 ⇒ A = =104 2 26
Now tan = =11
2 26
L
A
3. If cot = 6
8, then find other trigonometric ratios
Sol.
10
8
6
cot = 6
8 =
A
L
∴ A = 6, L = 8
Now K2 = A2+L2
= 62 + 82 = 36 + 64 = 100
K = 100 = 10
sin = =8
10
L
K, cosec =
=
1 10
sin 8
cos = =6
10
A
K, sec =
=
1 10
cos 6
tan = =8
6
L
A
4. If tan = a
b, find the value of
−
+
sin cos
sin s
a b
a bco
Sol. tan = a
b =
L
A
+2 2a b
a
b
K = +2 2a b
sin =+2 2
L a
K a b
cos = =+2 2
A b
K a b
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Now,
−
+
sin cos
sin cos
a b
a b
=
− + +
+ + +
2 2 2 2
2 2 2 2
a ba b
a b a ba b
a ba b a b
=
−
+
+
+
2 2
2 2
2 2
2 2
a b
a b
a b
a b
= −
+
2 2
2 2
a b
a b
2nd Method
tan = a
b
Now,
−
+
sin cos
sin cos
a b
a b
=
−
+
sin cos
cos cossin cos
cos cos
a b
a b
=
−
+
tan
tan
a b
a b =
−
+
aa b
ba
a bb
= −
+
2 2
2 2
a b
a b
5. If sin = a
bthen the value of sec - cos is (where 0ᵒ< <
90ᵒ)
Sol. sin = a
b =
L
K
b
a
−2 2b a
Base (A) = −2 2b a
cos = −
=2 2A b a
K b
sec =
1
cos=
−2 2
b
b a
sec - cos = −2 2
b
b a
−
−2 2b a
b
= ( )− −
−
2 2 2
2 2
b b a
b b a =
−
2
2 2
a
b b a
6. if
+
−
tan cot
tan cot = 2 then the value of sin =
Sol.
+
−
tan cot
tan cot = 2
Using componendo dividendo
( )( )
+ + −
+ − −
tan cot tan cot
tan cot tan cot =
+
−
2 1
2 1 ⇒
2tan
2cot = 3
⇒
tan
1cot
= 3 ⇒ tan2 = 3 ⇒ tan = =3
1
L
A
2
3 a
1
K = +3 1 = 2
sin = L
K =
3
2
7. For the right angle ∆ABC, Find the values of all six
trigonometric ratios of acute angles A and B.
5
3
4
B
A C
Sol. For angle A, Base = 4, perpendicular = 3, hypotenuse = 5
sin = =Perpendicular 3
hypotenuse 5
cos A = 4
=Hypotenuse 5
Base
tan = Perpendicular 3
=Base 4
cot A = 1
tan A=
4
3
sec A = 1
cos A=
5
4
cosec A = 1
sin A =
5
3
for angle B, Base (A) = 3, Perpendicular (L) = 4,
hypotenuse (K) = 5
sin B = 4
5, cos B =
3,
5
tanB = 4
3, cot B =
3
4
sec B = 5
3, cosec B =
5
4
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11
1. Value of cot is
√5
1
2
(a) 1
2 (b)
5
2 (c) 2 (d)
1
5
2. Value of tan A is
√5
5
13
12
(a) 5
12 (b)
2
5 (c)
5
13 (d)
12
13
3. Values of sin , tan are
41
40
9
(a) 9 41
,40 9
(b) 40 40
,41 9
(c) 9 40
,41 410
(d) 40 41
,9 40
4. Find values of cosA, cosecA –
41
A C
15
17
8
(a) 8 17
,17 15
(b) 8 15
,17 17
(c) 17 8
,8 15
(d) 17 15
,15 8
5. Find values of tanA, cotB
11
60
A
B C
(a) 11
60,
11
60 (b)
11
60,60
11 (c)
60 61,
61 60 (d)
60
11,
60
11
6. Find the value of sinB
3 A
6
C
B D
(a) 2
3 (b)
2
13 (c)
4
5 (d)
3
13
7. In ∆ABC, ∠B = 90ᵒ c = 12 cm and a = 9cm then cos C is
(a) 3
5 (b)
3
4 (c)
3
4 (d)
4
5
8. In ∆ABC, ∠B = 90ᵒ and AB : BC = 2 :1. Then value of
(cosA + tanC) is
(a) 2 + 5 (b) ( )2 1 5
5
+ (c)
2 5
5
+ (d) 1 + 5
9. if sin = 2 2
2 2
a b
a b
−
+ then find cot
(a) 2 2
2 2
a b
a b
+
− (b)
2 2
2ab
a b+ (c)
2 2
2ab
a b− (d)
2 2
3ab
a b+
Exercise - 1
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12
10. If cot = 40
find values of cosec and sec
(a) 41
40,
41
9 (b)
41
9,41
40 (c)
40
41
40
9 (d)
40
9,40
41
11. If tan = 4
3, then the value of
3sin 2cos
3sin 2cos
+
−is
(a) 0.5 (b) -0.5 (c) 3.0 (d) -3.0
12. If 5tan = 4, then 5sin 3cos
5sin 2cos
−
+ is equal to
(a) 2
3 (b)
1
4 (c)
1
6 (d)
1
3
13. If sec tan 5
sec tan 3
+=
− then sin?
(a) 1
4 (b)
1
3 (c)
2
3 (d)
3
4
14. sin cos
sin cos
+
− = 3 then the value of sin2 is
(a) 4
5 (b)
2
5 (c)
1
5 (d)
3
5
15. if tan = 1
11 and 0ᵒ<<
2
then the value of
2 2
2 2
cos sec
cos sec
ec
ec
−
+ is
(a) 3
4 (b)
4
5 (c)
5
6 (d)
6
7
16. sin cos
sin cos
+
−=
5
4then the value of
2
2
1 tan
2 tan
+=?
(a) 25
16 (b)
41
81 (c)
41
80 (d)
40
41
17. if secA = 5
4 then the value of
cot 1
cot 1
A
A
+
− is
(a) 1
4 (b)
2
7 (c)
3
7 (d) 7
18. if sin = 2 2
a
a b+ 0ᵒ < < 90ᵒ, Find the value of cos
and tan
(a) 2 2
,b a
ba b+ (b)
2 2,
+
b b
aa b
(c) 2 2
2 2,
+
+
b a b
aa b (d)
2 2
2 2,
+
+
a b a
a a b
19. if cosA = 5
13, then
cos
cos cos
ecA
A ecA+ ?
(a) 169
229 (b)
229
169 (c) 169 (d) 229
20. If cos = 3
5 , then the value of sin cos cot ?
(a) 16
25 (b)
9
16 (c)
9
25 (d)
4
3
21. If sec tan 51
2sec tan 79
+=
− then the value of the sec - tan
(a) 91
144 (b)
39
72 (c)
65
144 (d)
35
72
22. If sin = a
b, find sec + tan in terms of a and b
(a) b a
b a
+
− (b)
a b
a b
+
− (c)
b a
b a
+
− (d)
b a
a b
+
−
23. If cos = 3
5, Find the value of
sin cot
2
cot
−
(a) 160
3 (b) 2 (c)
3
160 (d) 3
24. if tan = 20
21, then
1 sin cos
1 sin cos
− +
+ +=?
(a) 3
7 (b)
7
3 (c)
20
7 (d)
7
20
25. if is an acute angle such that cos = 3
5 then
2
sin tan 1
2 tan 1
−
−
(a) 16
625 (b)
1
36 (c)
3
115 (d)
160
3
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13
1. If angle C of triangle ABC is 90ᵒ, then tanA + tanB =?
(Where a,b,c are sides opposite to angles A, B, C
respectively)
(a) ab
c (b)
2 2
2
a b
c
− (c)
2c
ab (d)
2
2 2
c
a b−
2. tanA = 2 1− then cosA= ?
(a) 4 2 2− (b) 1
4 2 2−
(c) 4 (d) 1
4 2−
3. Which of the following is true for a right angled
triangle?
(a) sin2 + cos2 = 1 (b) sec2 - tan2 = 1
(c) cosec2 - cot2 = 1 (d) All of these
4. In a right angled triangle XYZ right angled at Y, XY = 2√6
cm and XZ – YZ = 2, then sec X + tan X is
(a) 1
6 (b) 6 (c) 2 6 (d)
6
2
5. If cosA = 2 2
3, then cosA cosecA + tanA secA = ?
(a) 16 2 3
8
+ (b)
7 5 3
8
+
(c) 16 2 5
7
+ (d)
16 7 5
7
+
6. In ∆PQR, right angled at Q, PR+QR = 25cm and PQ = 5
cm determine the values of sinP, cosP and tanP.
(a) 13 5 5
, ,12 13 12
(b) 12 5 12
, ,13 13 5
(c) 13 13 12
, ,12 5 5
(d) 12 13 13
, ,13 5 12
7. If 3cos = 1 find the value of 2 26sin tan
4cos
+
(a) 10 (b) 13 (c) 0 (d) 9
8. If secA = 5
4 then
3
3
3sin 4sin
4cos 3cos
−
−
A A
A A
(a) 44
117 (b)
117
44
− (c)
217
22 (d)
22
217
−
9. If is an acute angle and tan2 = 8
7 then the value of
( )( )
( )( )
1 sin 1 sin
1 cos 1 cos
+ −
+ −
(a) 7
8 (b)
8
7 (c)
7
4 (d)
64
49
10. In triangle ABC, right angled at C and AB = 29cm, BC =
21cm, ∠ABC = then find out value of cos2 + sin2
(a) 41
841 (b)
3
2 (c)
801
841 (d) 1
11. In a right angled triangle ABC, right angled at B, tan A =
1 then value of 2sinA cos A=?
(a) 1
2 2 (b) 1
(c) 2 2 (d) none of these
Exercise - 2
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14
Answer key
1 c 2 b 3 b 4 a 5 d
6 b 7 a 8 b 9 c 10 a
11 c 12 c 13 a 14 a 15 c
16 b 17 d 18 a 19 a 20 c
21 c 22 a 23 c 24 a 25 c
1. (c)
Here L = 1, A = 2, K = √5
cot = ( )( )
Base
perpendicular
A
L=
2
1 = 2
2. (b)
Here
L = 12, A = 5, K = 13
tanA = Perpendicular(L)
Base(A)=
12
5
[Note:- the side opposite to angle is called
perpendicular and adjacent to angle is called base]
3. (b)
Here
L = 40, A = 9, K = 41
sin = =40
41
L
K
tan = =40
9
L
A
4. (a)
Here
L = 15, A = 8, K = 17
cosA = =8
17
A
K
cosecA = =17
15
K
L
5. (d)
For angle A
L = 60, A = 11, K = 61
∴tanA = L
A =
0
11
For angle B
L = 11, A = 60, k = 61
cot B = A
L =
60
11
6. (b)
∆BDC is right angle triangle
For angle B
A = 6, L = 4
K = 2 26 4+ = 52 = 2 13
sinB = 4 2
2 13 13
L
K= =
7. (a)
b
12
9
A
C B
(Note - c is the side opposite to angle C and a and b are
also sides opposite to angles ∠A and ∠B respectively.)
b = 2 212 9+ = 15
here L = 12 cm A = 9cm, K = 15 cm
cos C = 9 3
15 5
A
K= =
8. (b)
Let AB and BC be 2 and 1 respectively.
Then AC = 2 22 1 5+ =
Solutions
Exercise - 1
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15
5
1
2
C
A B
For angle A,
L = 1, A = 2, K = 5
cos A = 2
5
A
K=
For angle C,
L =2, A = 1 K = 5
tanC = 2
1 = 2
cosA + tanC = 2
5+ 2 =
( )2 1 5
5
+
9. (c)
2 2a b+
2 2a b−
A
C B Here
L = 2 2a b− , k = 2 2a b+
A = 2 2−K L = ( ) ( )2 2
2 2 2 2+ − −a b a b
= ( )4 4 2 2 4 4 2 22 2+ + − + −a b a b a b a b = 2ab
∴ cot = 2 2
2A ab
L a b=
−
10. (a)
41 40
9
A
C B
cot = 9
40
Here L = 40, A = 9
9, 40 and 41 are triplet
∴ K = 41
Now cosec = 41
40
k
l=
sec 41
9
k
A=
11. (c)
tan = 4
3
now 3sin 2cos
3sin 2cos
+
− =
sin cos3 2
cos cossin cos
3 2cos cos
+
−
= 3tan 2
3tan 2
+
− =
43 2
34
3 23
+
−
= 6
2 = 3
12. (c)
5 tan = 4 ⇒ tan = 4
5
5sin 3cos
5sin 2cos
−
+ =
5 tan 3
5 tan 2
−
+ =
45 3
54
5 25
−
+
= 1
6
13. (a)
sec tan 5
sec tan 3
+=
−
Using C & D
⇒ ( )
( )
sec tan sec tan 5 3
sec tan sec tan 5 3
+ + + +=
− − − − ⇒
2sec
2 tan
= 4
⇒
1
cossin
cos
⇒ 1
sin = 4 ⇒ sin =
1
4
Alternate:
sec tan 5
sec tan 3
+=
−
3(sec + tan) = 5(sec-tan) ⇒ 3sec + 3tan = 5sec-5tan
⇒ 2sec = 8 tan ⇒ 2 sin
8cos cos
= ⇒ sin =
2 1
8 4=
14. (a)
sin cos
sin cos
+
− = 3 ⇒ sin +cos = 3 sin - 3cos
⇒ 2sin = 4cos ⇒ sin 4
cos 2
= = 2 ⇒ tan = 2
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16
2
1
A
C B
here L = 2, A = 1, K = 5
sin = 2
5
L
K=
sin2 =
2
2 4
55
=
15. (c)
1
A
C B
tan = 1
11
L = 1, A = 11
K = 11 1+ = 12 = 2 3
cosec = 2 3
1 = 2 3
sec = 2 3
11
w 2 2
2 2
cos sec
cos sec
ec
ece
−
+
=
( )
( )
22
22
2 32 3
11
2 32 3
11
−
+
=
1212
120 51112 144 6
1211
−
= =
+
16. (b)
sin cos
sin cos
+
− =
5
4
⇒ 4(sin + cos) = 5(sin − cos)
⇒ sin = cos ⇒ tan = 9
Now
2
2
1 tan 1 81
2 812 tan
+ +=
=
82 41
2 81 81=
17. (d)
5
3
4
secA = 5
4 =
h
b
L = 2 25 4− = 3
∴ cot A = 4
3
cotA 1
cot 1A
+
−=
41
34
13
+
−
=
7
31
3
= 7
18. (a)
2 2a b+
a
b
Here L = a
K = 2 2a b+
A = 2 2 2a b a+ − = b
cos = 2 2
A b
K a b=
+
tan = L a
K b=
19. (a)
13
12
5
Base (A) 2 213 5− = 12
∴ cosec A = 13
12
K
L=
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17
cos
cos cos
ecA
A ecA+ =
13
125 13
13 12+
=
13
1260 169
156
+ =
13 156 169
229 12 229
=
20. (c)
sin cos cot = sin cos cos
sin
= cos2 =
23 9
5 25
=
21. (c)
sec tan
sec tan
+
−=
512
79 ⇒
sec tan 209
sec tan 79
+=
−
Using C & D
⇒ ( )( )
sec tan sec tan
sec tan sec tan
+ + −
+ − − =
209 79
209 79
+
−
⇒ 2sec 288
2 tan 130
= ⇒
1
144cossin 65
cos
= ⇒ sin = 65
144
22. (a)
b
a
2 2b a−
sin = a
b
Base (A) = 2 2b a−
sec = 2 2
=−
K b
A b a
tan = 2 2
L a
A b a=
−
sec + tan = 2 2 2 2
+− −
b a
b a b a
= 2 2
b a
b a
+
− =
+
+ −
b a
b a b a=
b a
b a
+
−=
b a
b a
+
−
23. (c)
5
4
3
cos = 3
5, sin =
4
5
cot = 3
4
Now,
4 3
sin cot 5 42 2
3cot
4
−−
= =
1
208
3
= 3
160
24. (a)
29
20
21
L = 20, A = 21
sin = 20
29 cos =
21
29
1 sin cos
1 sin cos
− +
+ + =
20 211
29 2920 21
129 29
− +
+ +
=
29 20 21
2929 20 21
29
− +
+ + =
30 3
70 7=
25. (c)
5
4
3
cos = 3
5
∴ sin = 4
5, tan =
4
3
Now
2
sin tan 1
2 tan 1
−
− =
4 41
5 316
2 19
−
−
=
1
1523
9
= 9 3
15 23 115=
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19
Answer key
1 c 2 b 3 d 4 b 5 a
6 b 7 a 8 b 9 a 10 d
11 b
1. (c)
c
b
a
A
B C
tanA = a
b
tanB = b
a
tanA + tanB = a b
b a+ =
2 2a b
ab
+ =
2c
ab
2. (b)
1
B
A C
tan = 2 1−
K = ( )2
22 1 1− + = 2 1 2 2 1+ − + = 4 2 2−
cosA = 1
4 2 2=
−
base
hypotenuse
3. (d)
Let ABC a right angled triangle right angled at B
A
B c
Here sin = AB
AC, cosec =
AC
AB
cos = BC
AC, sec =
AC
BC
tan = AB
BC, cot =
BC
AB
and AC2 = AB2 + BC2
sin2 + cos2 = 2 2
AB BC
AC AC
+
= 2 2 2
2 2
AB BC AC
AC AC
+= = 1 (true)
sec2 - tan2 = 2
2
AC
BC-
2
2
AB
BC =
2 2
2
AC AB
BC
−
= 2
2
BC
BC = 1 (true)
cosec2 - cot2 = 2
2
AC
AB-
2
2
BC
AB=
2 2
2
AC BC
AB
−
= 2
2
AB
AB = 1 (true)
4. (b)
2√6
Z
X Y
Let YZ be a cm then XZ = a + 2
Then in ∆ XYZ
(XZ)2 = (YZ)2 + (XY)2
⇒ (a+2)2 = a2 + (2√6)2
⇒ a2+4a+4 = a2+24 ⇒ 4a = 20 ⇒ a = 5
∴ YZ = 5cm, XZ = 7 cm
secX +tanX = 7 5
2 6 2 6+ =
126
2 6=
Exercise - 2
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20
5. (a)
3
1
2 2
B
A C
cosA = 2 2
3
BC = ( )2
23 2 2− = 1
cosecA = 3, secA = 1 3
cos 2 2A=
tanA = 1
2 2
cosA cosec A + tanA secA
= 2 2 1 3
33 2 2 2 2
+ = 2√2 +3
8 =
16 2 3
8
+
6. (b)
5
R
P Q
Let QR = x cm
Then PR = 25 – x cm
In ∆PQR
PR2 = PQ2+QR2
(25-x)2 = 52+x2 ⇒ 625+x2-50x = 25+x2
⇒ 50x = 600 ⇒ x = 12cm
∴ QR = 12cm, PR = 13cm
∴sinP = 12
13
cos P = 5
13
tan P = 12
5
7. (a)
3
2√2
1
cos = 1
3
sin = 2 2
3, tan = 2 2
2 26sin tan
4cos
+=
86 8
91
43
+
= 120 3
9 4
= 10
8. (b)
5
3
4
C
A B
BC = 2 25 4− = 3
sin A = 3
5
cos A = 1 4
sec 5A=
Now
3
3
3sin 4sin
4cos 3cos
A A
A A
−
− =
3
3
3 33 4
5 5
4 44 3
5 5
−
−
=
9 108
5 125256 12
125 5
−
−
= 225 108
256 300
−
−= -
117
44
9. (a)
tan2 = 8
7
tan = 2 2
7
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21
15 2√2
7
( )( )
( )( )
1 sin 1 sin
1 cos 1 cos
+ −
+ −
= 2
2
1 sin
1 cos
−
− [∵ (a+b)(a-b) = a2 – b2]
Here = sin = 2 2
15
L
K=
cos = 7
15
A
K=
=
2
2
2 21
15
71
15
−
+
=
81
157
115
−
−
=
7
7158 8
15
=
10. (d)
21
A
B C
AC = 2 2 2 229 21AB BC− = −
= ( )( )29 21 29 21− + = 8 50 400 = = 20 cm
Now
cos = 21
29
A
K=
sin = L
K =
20
29
cos2+sin2 = 2 2
21 20
29 29
+
= 441 400
841
+ = 1
Note : Value of sin2 + cos2 will be always 1
irrespective of the value of
11. (b)
C
A B
tan A = 1 = BC
AB
∴ BC = AB = 1
and AC 2 2BC AB= + = 1 1 2+ =
sin A = 1
2
C
AC
=
cos A = 1
2
AB
AC=
2sin A cos A = 2×1 1
2 2 = 1
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0ᵒ 30ᵒ 45ᵒ 60ᵒ 90ᵒ
sin 0 1
2
1
2 3
2 1
cos 1 3
2
1
2
1
2 0
tan 0 1
3 1 3
Not
define
cot Not
define 3 1
1
3 0
sec 1 2
3 2 2
Not
define
cosec Not
define 2 2
2
3 0
Examples 1. Find the value of tan260ᵒ cosec30ᵒ tan45ᵒ
Sol. tan260ᵒ cosece30ᵒtan45ᵒ
= ( )2
3 ×2×1 = 6
2. Find the value of (cos0ᵒ + sin45ᵒ + sin30ᵒ)(sin90ᵒ-
cos45ᵒ+cos60ᵒ)
Sol. (cos0ᵒ+sin45ᵒ+sin30ᵒ)(sin90ᵒ-cos45ᵒ+cos60ᵒ)
= 1 1 1 1
1 12 22 2
+ + − +
= 2 2 1 2 2 1
2 2
+ + − +
= 3 2 3 2
2 2
+ −
= 9 2 7
4 4
−=
3. Value of cosec30ᵒ + cot45ᵒ?
Sol. cosec30ᵒ + cot45ᵒ
= 2 + 1 = 3
4. If = 45ᵒ then 2sin cos =?
Sol. = 2sin45ᵒ cos45ᵒ
= 2×1 1
2 2 = 1
5. sin30 tan 45 sin60 cos30
sin 45 sec60 cot 45 sin90
+ − −
Sol.
1 3 3
12 2 21 2 1 1
2
+ − −
= 1 1 3 3
2 2 22+ − −
= 1 1
322
+ − = 2 1 2 3
2
+ −
6. Value of 2 2 25sin 30 cos 45 4 tan 30
2sin30 cos30 tan 45
+ −
+
Sol. 2 2 25sin 30 cos 45 4 tan 30
2sin30 cos30 tan 45
+ −
+
=
2221 1 1
5 42 2 3
1 32 1
2 2
+ −
+
=
1 1 15 4
4 2 3
31
2
+ −
+
=
5 1 4
4 2 3
3 1
2
+ −
+ =
15 6 16
12
3 1
2
+ −
+
= 5 2
12 3 1
+ =
( )5
6 3 1+
7. If 3tan22x = cos60ᵒ+sin45ᵒcos45ᵒ then x =?
Sol. 3tan22x = cos60ᵒ+sin45ᵒcos45ᵒ
3tan22x = 1 1 1
2 2 2+
Ratios of Standard Angles 3
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⇒ 3tan22x = 1 1
2 2+ = 1
⇒ tan22x = 1
3 ⇒ tan2x =
1
3= tan30ᵒ
⇒ 2x = 30ᵒ ⇒ x = 15ᵒ
8. If sin(x-y) =1
2 and cos(x+y)=
1
2 then value of x is
Sol. sin(x-y) = 1
2
⇒ sin(x-y) = sin30ᵒ
⇒ (x-y) = 30ᵒ ...................... (i)
cos(x+y) = 1
2
⇒ cos (x+y) = cos (60ᵒ)
⇒ (x+y) = 60ᵒ .................. (ii)
(i) + (ii)
2x = 90ᵒ
X = 45ᵒ
Y = 60 - 45ᵒ = 15ᵒ
9. In ∆ABC, right angled at B, AB = 3cm and AC = 6cm find
∠BAC and ∠ACB
Sol.
6
3
A
C B
BC = 2 26 3 3 3− =
sinC = 3 1
6 2
AB
AC= = = sin30ᵒ
∴ ∠ACB = 30ᵒ
∠BAC = 90 – 30ᵒ = 60ᵒ
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24
1. cos60ᵒ cos30ᵒ - sin60ᵒsin30ᵒ = ?
(a) 1 (b) 0 (c) 1
2 (d) -1
2. tan230ᵒ+tan260ᵒ+tan245ᵒ
(a) 13
3 (b)
1
12 (c)
13
2 (d)
1
13
3. cos30 sin60
1 cos60 sin30
+
+ + =?
(a) 3 (b) 3
4 (c)
3
2 (d) 1
4. 4cot260ᵒ + sec230ᵒ-sin245ᵒ =?
(a) 5
2 (b)
17
6 (c)
13
6 (d)
19
6
5. 2
2 tan30
1 tan 30
+ =?
(a) sin60ᵒ (b) cos60ᵒ (c) tan60ᵒ (d) sin30ᵒ
6. cos60ᵒcos45ᵒ + sin60ᵒsin45ᵒ
(a) 2
3 1+ (b)
3 1
3 1
−
+ (c)
3 1
2 2
− (d)
2 2
3 1+
7. sin230ᵒ+sin245ᵒ+sin260ᵒ+sin290ᵒ
(a) 2
5 (b)
3
5 (c)
5
2 (d)
5
3
8. if 3 sinx = cosx then x = ?
(a) 30ᵒ (b) 45ᵒ (c) 60ᵒ (d) 0ᵒ
9. Find value of cot22 2 2cot cot cot
2 3 4 6
+ + +
(a) 15
4 (b)
17
4 (c)
13
3 (d)
14
3
10. cosec290ᵒ + cosce2 60ᵒ + cosece2 45ᵒ + cosce2 30ᵒ
(a) 23
4 (b)
17
6 (c)
13
3 (d)
25
3
11. tan60ᵒ cosece245ᵒ + sec260ᵒtan245ᵒ
(a) 4+ 2 3 (b) 4 - 2 3 (c) 2 - 3 (d) 2 + 3
12. 2sin230ᵒtan260ᵒ - 3cos260ᵒsec230ᵒ
(a) 1
2 (b)
2
3 (c)
3
4 (d)
4
3
13. tan 45
cos 30ec
+
sec60 5sin90
cot 45 2cos0
−
=?
(a) 0 (b) 2
5 (c) 1 (d)
1
5
14. 4 4 2 2 22 cos sin tan cot 3sec
3 6 3 4 6
+ − + +
=?
(a) 1
4 (b)
2
3 (c)
4
7 (d)
4
3
15. sin230ᵒcos245ᵒ + 4tan230ᵒ +1
2sin290ᵒ - 2cos290ᵒ +
1
24
cos0ᵒ
(a) 1 (b) -1 (c) 0 (d) 2
16. 2 2 2
2 2
cos 60 4sec 30 tan 45
sin 30 cos 30
+ −
+
(a) 64
3 (b)
55
12 (c)
67
12 (d)
67
10
17. (cosec245ᵒsec230ᵒ)(sin230ᵒ+4cot345ᵒ-sec260ᵒ)
(a) 1
2 (b) (c)
2
3 (d)
3
2
18. tan30ᵒsec45ᵒ + tan60ᵒsec30ᵒ
(a) 2 2 3
3
+ (b)
3 2 2
3
+
(c) 2 2 3
2 3
− (d)
3 2 2
2 3
−
Exercise - 1
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19. sin30 cos45
tan 60
=?
(a) 6
12 (b)
2
3 (c) 2 3 (d)
2
3
20. If sinx = 3 cosx then cosec2
x =?
(a) 2
3 (b) 2 (c) 2 (d) Not define
21. 2sin2
x = 1 then cosx =?
(a) 3
2 (b)
1
2 (c)
3
2 (d)
1
2
22. 2cos2
2
x = 1 than tanx =?
(a) 3 (b) 1
3 (c) 1 (d) not defined
23.
2 2 2 2
2 2 2 2
cos 45 cos 60 tan 30 sin 30
sin 60 sin 45 cot 45 cot 30
+ − −
(a) 1
4 (b)
3
4 (c)
1
2 (d)
5
4
24. sin3
3
cot
6
- 2sec2
4
+3 cos
3
tan
4
- tan2
3
(a) 35
8 (b)
35
8− (c)
17
4− (d)
11
8
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1. if sin(A-B) = sinA cosB – cosA sinB then find value of
sin15ᵒ
(a) 3 2
2
− (b)
3 1
2 2
+ (c)
3 2
2 2
− (d)
3 1
2 2
−
2. if 2sin2x+cos245ᵒ = tan45ᵒ & 0ᵒ < x < 90ᵒ then sec2x = ?
(a) 1
2 (b)
2
3 (c) 2 (d) 2
3. 2 2
2 2
cos 30 sec 45
8cos 45 sin 60
x ec
= tan2 60ᵒ - tan2 30ᵒ, then x =?
(a) 0 (b) 1 (c) 2 (d) 2
4. x sin60ᵒtan30ᵒ - tan245ᵒ = cosec60ᵒcot30ᵒ - sec245ᵒ
then x = ?
(a) 2 (b) -2 (c) 6 (d) -4
5.
2
2 2
2
2 tan6 sec sec 0 sec
4 31 tan
6
x
+ + =
−
then x = ?
(a) 2 (b) 1 (c) 0 (d) -1
6. if x cos45ᵒ = y sec60ᵒthen 4
4
x
y
(a) 43 (b) 63 (c) 23 (d) 83
7. If tan (A+B) = 3 and tan (A-B) =1
3; 0ᵒ < A+B ≤ 90ᵒ;
A>B then find A and B?
(a) ,3 4
(b) ,
3 6
(c) ,
2 6
(d) ,
4 12
8. If and are positive acute angles sin (4−) = 1 and
cos (2+) =1
2, then the value of cos (+) is
(a) 0 (b) 1 (c) 3
2 (d)
1
2
9. If sin + cos = 2 (0ᵒ ≤ , ≤ 90ᵒ) then tan2
3
+
= ?
(a) 1
3 (b) 3 (c) 1 (d) 1
10. In an acute angled triangle ABC, if tan (A+B-C) = 1, sec
(B+C-A) = 2 and sin (C+A-B) = 1
2, find the value of A, B
and C
(a) 1 1
77 , 45 , 222 2
A B C
= = =
(b) 1 1
52 , 55 ,C 372 2
A B
= = =
(c) 1 1
60 , 52 ,C 672 2
A B
= = =
(d) 1 1
37 , 52 , C 452 2
= = =A B
11. If tan (+) = 3 and sec (1-2) = 2
3 then the value of
sin21 + tan32 is equal to (0 < − < + ᵒ)
(a) 0 (b) 3 (c) 2 (d) 1
12. If tanA = 1, tan B = 3 , then 1 1
1 cos1 sin
2
++
−BA
=?
(a) 2 1
2 3
− (b)
3 2 2
2 1
+
+ (c)
3 1
2 2
− (d)
2 3
2 1+
13. If 2
2 2
cos
cot cos
− = 3 and 0ᵒ < < 90ᵒ then the value of
is : -
(a) 30ᵒ (b) 45ᵒ (c) 60ᵒ (d) None of these
14. If sin(A+B)= 1
2 and sin (A-B) =
1
2 then (cos2 B – cos2 A
) = ?
(a) 1
2 (b) 1 (c) 0 (d) 2
15. Which of the following is true (0ᵒ < < 90ᵒ)
(a) if 2 cos 3 = 1 then = 20ᵒ
(b) if 2 sin 2 = 3 then = 9ᵒ
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(c) if tan 5 = 1 then = 30ᵒ
(d) All of these
16. Find an acute angle , when sin cos 3 1
sin cos 3 1
− −=
+ +
(a) 60ᵒ (b) 90ᵒ (c) 45ᵒ (d) 30ᵒ
17 If sin2x = sin60ᵒ cos30ᵒ - cos60ᵒ sin30ᵒ then find the
value of x?
(a) 55ᵒ (b) 100ᵒ (c) 65ᵒ (d) 15ᵒ
18. If each of is a positive acute angle such that cosec
(+−) = 2
3 sec (+−) = 2 and cot (+−) = 1, find
the value of , ?
(a) 1 1
33 , 45 , 522 2
= = =
(b) 1
82 , 602
= = =
(c) 1
52 , 602
= = =
(d) 1
52 , 602
= = =
19. if cosec = 2 then 2 2
2 2
2sin 3cot
4 tan cos
+
−
(a) 4
5 (b)
3
4 (c)
8
7 (d)
1
7
20. If + = 90ᵒ, and = 2 then cos2 + sin2 + tan2 +
cot2 =
(a) 15 (b) 15
4 (c)
13
3 (d)
17
4
21. if 2
2
tan 30
1 cot 60
x x−
+ = sin230ᵒ+4cot245ᵒ- cosec230ᵒ the
value of x2 is:
(a) 1
4 (b)
1
3 (c)
1
2 (d)
1
5
22. if = 60ᵒ then 1
1 sin 1 sin
+
+ − = ?
(a) 8 (b) 0 (c) 4 (d) 2
23. If cosecA = 2 find the value of 1 sin
tan 1 cos
A
A A+
+
(a) 1 (b) 0 (c) -1 (d) 2
24. Find acute angle A and B. if sin (A+2B) = 3
2 and cos
(A+4B) = 0, A > B
(a) A = 30ᵒ, B = 15ᵒ (b) A = 30ᵒ, B = 45ᵒ
(c) A = 60ᵒ, B = 15ᵒ (d) A = 55ᵒ, B = 30ᵒ
25. If + =2
and cos =
3
2, then value of sec =?
(a) 1 (b) 2
3 (c) 2 (d) 2
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Answer key 1 b 2 a 3 c 4 c 5 c
6 a 7 c 8 a 9 c 10 d
11 a 12 a 13 a 14 a 15 d
16 b 17 c 18 a 19 a 20 c
21 b 22 d 23 b 24 b 25 c
1. (b)
cos60ᵒ cos30ᵒ - sin60ᵒsin30ᵒ
= 1
2×
3 3 1
2 2 2− = 0
2. (a)
tan230ᵒ+tan260ᵒ+tan245ᵒ
= ( )2
213 1
3
+ +
=
1
3+3+1 =
13
3
3. (c)
cos30 sin60
1 cos60 sin30
+
+ + =
3 3
32 21 1 2
12 2
+
=
+ +
4. (c)
4cot260ᵒ + sec230ᵒ-sin245ᵒ
= 4
2 2 2
1 2 1
3 3 2
+ −
= 4 4 1 8 1 16 3 13
3 3 2 3 2 6 6
−+ − = − = =
5. (a)
2
2 tan30
1 tan 30
+ =
2
12
3
11
3
+
=
2
3
11
3+
=
2
3
4
3
= 2 3
4 3
=
3
2 = sin60ᵒ
6. (c)
cos60ᵒcos45ᵒ + sin60ᵒsin45ᵒ
= 1 1 3 1
2 22 2 + =
3 1
2 2
+
7. (c)
sin230ᵒ+sin245ᵒ+sin260ᵒ+sin290ᵒ
=
222
21 1 31
2 22
+ + +
= 1 1 3
14 2 4
+ + + = 1 5
1 12 2
+ + =
8. (a)
3 sinx = cosx ⇒ sin 1
cos 3
x
x= ⇒ tanx =
1
3
tanx = tan30ᵒ
X = 30ᵒ
9. (c)
cot22 2 2cot cot cot
2 3 4 6
+ + +
= ( ) ( )2
2210 1 3
3
+ + +
= 1 13
1 33 3
+ + =
10. (d)
cosec290ᵒ + cosce2 60ᵒ + cosece2 45ᵒ + cosce2 30ᵒ
= ( ) ( )2
2 22 21 2 2
3
+ + +
= 1+
4
3+2+4 =
25
3
11. (a)
tan60ᵒ cosece245ᵒ + sec260ᵒtan245ᵒ
= ( ) ( )2 2
3 2 2 1+ = 2 3 4+
12. (a)
Solutions
Exercise - 1
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2sin230ᵒtan260ᵒ - 3cos260ᵒsec230ᵒ
( )2 2
21 1 22 3 3
2 2 3
−
= 1 1 4
2 3 34 4 3
− = 3 1
12 2
− =
13. (a)
tan 45
cos 30ec
+
sec60 5sin90
cot 45 2cos0
−
= 1 2 5
2 1 2+ − =
5 50
2 2− =
14. (a)
4 4 2 2 22 cos sin tan cot 3sec
3 6 3 4 6
+ − + +
= ( ) ( )24 4
2 21 1 22 3 1 3
2 2 3
+ − + +
= 1 1 4
2 3 1 316 16 3
+ − + +
=
1 14 4
4 4− + =
15. (d)
sin230ᵒ cos245ᵒ + 4tan230ᵒ + 1
2sin290ᵒ - 2cos290ᵒ +
1
24
cos0ᵒ
= ( ) ( ) ( )22
1 1 1 1 14 1 2 0 1
2 2 242 3
+ + − +
=1 1 1 1 1
4 04 2 3 2 24
+ + − +
= 1 4 1 1
8 3 2 24+ + +
3 32 12 1
24
+ + += = 2
16. (b)
2 2 2
2 2
cos 60 4sec 30 tan 45
sin 30 cos 30
+ −
+
=
22
22
1 24 1
2 3
1 3
2 2
+ −
+
=
1 161
3 64 12 554 3
1 12 12
+ −+ −
= =
17. (c)
(cosec245ᵒsec230ᵒ)(sin230ᵒ+4cot345ᵒ-sec260ᵒ)
= ( ) ( ) ( )2 2
2 3 22 12 4 1 2
23
+ −
= 8 1 2
4 43 4 3
+ − =
18. (a)
tan30ᵒsec45ᵒ + tan60ᵒsec30ᵒ
= 1 2
2 33 3
+ = 2 2 2 3
23 3
++ =
19. (a)
sin30 cos45
tan 60
=
1 1
12 2
3 2 6
= = 6
12
20. (c)
sinx = 3 cosx ⇒ sin
3cos
x
x= ⇒ tanx = 3 ⇒ x = 60ᵒ
Now cosec 60
2
= cosec30ᵒ = 2
21. (b)
2sin2
x = 1 ⇒ sin
2
x =
1
2
⇒ sin2
x = sin30ᵒ ⇒
2
x = 30ᵒ, x = 60ᵒ
∴ cos60ᵒ = 1
2
22. (d)
2cos2
2
x = 1 ⇒ cos2
1
2 2
x=
⇒ cos2
x =
1
2 = cos 45ᵒ ⇒
2
x = 45ᵒ
X = 90ᵒ
Now tan90ᵒ = not defined
23. (b) 2 2 2 2
2 2 2 2
cos 45 cos 60 tan 30 sin 30
sin 60 sin 45 cot 45 cot 30
+ − −
= ( ) ( )
22 2 2
2 2 2 2
11 1 1
2 232
113 3
22
+ − −
= 1 1 2 1 1
2 3 4 1 3 4 3
+ − −
= 2 1 1 1
3 2 3 12+ − − =
8 6 4 1 9 3
12 12 4
+ − −= =
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24. (b)
( ) ( )3
2 23 13 2 2 3 1 3
2 2
− + −
= 3 3 3
3 2 2 38 2
− + − = 9 3
4 38 2
− + −
= 9 3
78 2
+ − = 9 12 56 35
8 8
+ −= −
25. (c)
cos =3
2⇒ = 0ᵒ ⇒ = − = ᵒ
sec = sec60ᵒ = 2
Answer key 1 d 2 c 3 b 4 a 5 a
6 a 7 d 8 d 9 b 10 d
11 c 12 b 13 c 14 a 15 d
16 a 17 d 18 c 19 c 20 c
21 a 22 a 23 d 24 a 25 c
1. (d)
15ᵒ = 45ᵒ - 30ᵒ
Sin (45ᵒ-30ᵒ) = sin45ᵒcos30ᵒ-cos45ᵒsin30ᵒ
= 1 3 1 1
2 22 2− =
3 1
2 2
−
2. (c)
2sin2x +
2
1
2
= 1 ⇒ 2sin2x + 1
2 = 1
⇒ 2sin2x = 1
2 ⇒ sin2x =
1
4 ⇒ sinx =
1
2
x = 300 (0ᵒ < x < 90ᵒ)
sec 2x = sec 60ᵒ = 2
3. (b)
2 2
2 2
cos 30 sec 45
8cos 45 sin 60
x ec
= tan2 60ᵒ - tan2 30ᵒ
⇒( ) ( )
( )22 2
2
22
2 2 13
31 38
22
= −
x
⇒ 4 2 13
1 3 38
2 4
x = −
⇒ 8 8
3 3
x= ⇒ x = 1
4. (a)
3 1
2 3x - 1 = ( )
223 2
3 −
⇒ 12
x− = 2 – 2 = 0 ⇒
2
x = 1, x = 2
5. (a)
2
2 2
2
2 tan6 sec sec 0 sec
4 31 tan
6
x
+ + =
−
Exercise - 2
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31
⇒
12
31
13
−
+2+1 = 2x ⇒ 1+3 = 2x ⇒ x = 2
6. (a)
x cos45ᵒ = y sec60ᵒ
x ×1
2 = y×2 ⇒
x
y = 2 2
⇒ ( ) ( )4
4 442 2 2 2
x
y
= =
= 24×22= 43
7. (d)
tan (A+B) = 3 = tan60ᵒ
(A+B) = 60ᵒ =3
....... (i)
tan (A-B) = 1
3 = tan30ᵒ
(A-B) = 30ᵒ =6
...... (ii)
(i) + (ii)
2A = 3 6 2
+ = ⇒ A =
4
And B = 3 4 12
− =
8. (d)
sin (4−) = 1 = sin90ᵒ
4− = 90ᵒ ....... (i)
cos (2+) = 1
2 = cos60ᵒ
2+ = 60ᵒ ….. (ii)
(i) + (ii)
= 150ᵒ ⇒ = 25ᵒ
And = 60ᵒ - 2×25ᵒ = 10ᵒ
Now
cos (+) = cos (25ᵒ+20ᵒ) = cos 45ᵒ = 1
2
9. (b)
sin+ cos = 2 possible iff sin = cos = 1
∴ = 90ᵒ, = 0ᵒ
Now tan 2
3
+
= tan2 90 0
3
+
= tan60ᵒ = 3
10. (d)
tan (A+B-C)=1 = tan45ᵒ
⇒ A+B-C = 45ᵒ ......... (i)
sec (B+C-A) = 2 = sec60ᵒ
⇒ B+C-A = 60ᵒ ........ (ii)
sin(C+A-B) = 1
2 = sin 30ᵒ
⇒ C+A-B = 30 ....... (iii)
(i)+(ii) 2B = 105ᵒ ⇒ B = 521
2
(ii)+(iii) 2C = 90ᵒ ⇒ C = 45ᵒ
(i)+(iii) 2A = 75ᵒ ⇒ A = 371
2
11. (c)
tan (+) = 3 = tan 60ᵒ
⇒ + = ᵒ (i)
sec (1−)= 2
3
⇒ − = 30ᵒ ....... (ii)
(i)+(ii), 2 = 90ᵒ
⇒ = 45ᵒ
And = 60ᵒ - 45ᵒ = 15ᵒ
Now
sin 21 + tan 3 = sin 90ᵒ+tan45ᵒ = 1+1 = 2
12. (b)
tanA = 1 A = 45ᵒ
tanB = 3 , B = 60ᵒ
1 1 1 1
1 cos 1 cos 45 1 sin 301 sin
2
BA+ = +
+ + − −
= 1 1
1 11 1
22
+
+ −
= 2 2 2 2 2
22 1 2 1
+ ++ =
+ +=
3 2 2
2 1
+
+
13. (c)
2
2 2
cos
cot cos
− = 3
⇒ cos2 = 3cot2 - 3cos2 ⇒ 4cos2 = 3 cot2
⇒ 4 cos2 = 32
2
cos
sin
⇒ sin2 =
3
4
⇒ sin = 3
2 ⇒ = 60ᵒ
14. (a)
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sin(A+B) = 1
2 = sin 45ᵒ
⇒ (A+B) = 45ᵒ .............. (i)
sin (A-B) = 1
2 = sin 45ᵒ
⇒ A – B = 45ᵒ ……(ii)
(i) + (ii)
2A = 90ᵒ ⇒ A = 45ᵒ
and B = 0
now cos2B – cos2A = cos20ᵒ - cos245ᵒ
= 1 -
2
1 1 11
2 22
= − =
15. (d)
(a) cos3 = 1
2 = cos60ᵒ
⇒ 3 = 60ᵒ, = 20ᵒ (true)
(b) sin2 = 3
2 = sin60ᵒ
⇒ 2 = 60ᵒ, = 30ᵒ (true)
(c) tan5 = 1 = tan 45ᵒ
⇒ 5 = 45ᵒ, = 9ᵒ (true)
16. (a)
sin cos 3 1
sin cos 3 1
− −=
+ +
By using C & D
( ) ( )( ) ( )
( )( ) ( )
3 1 3 1sin cos sin cos
sin cos sin cos 3 1 3 1
− + +− + +=
− − + + − −
⇒ 2sin 2 3
2cos 2
= ⇒ tan = 3 = tan 60ᵒ ⇒ = 60ᵒ
Alternate:
sin cos
sin cos
−
+ =
3 1
3 1
−
+
=
3 1
2 2
3 1
2 2
−
+
= sin 60 cos60
sin 60 cos60
−
+
∴ = 60ᵒ
17. (d)
sin2x = sin60ᵒ cos30ᵒ - cos60ᵒsin30ᵒ
sin2x = 3 3 1 1
2 2 2 2 − =
3 1
4 4−
⇒ sin2x = 1
2 ⇒ 2x = 30ᵒ
x = 15ᵒ
18. (c)
cosec (+−) = 2
3
+− = 60ᵒ ....... (i)
sec (+−)
+− = 60ᵒ ….... (ii)
cot (+−) = 1
⇒ +− = 45ᵒ ........ (iii)
(i) +(ii) 2 = 120ᵒ ⇒ = 60ᵒ
(ii)+(iii) 2 = 105ᵒ = =521
2
(i) +(iii) 2 = 105ᵒ ⇒ = 521
2
19. (c)
cosec = 2 = cosec45ᵒ
⇒ = 45ᵒ
Now 2 2
2 2
2sin 3cot
4 tan cos
+
−
=
( )
( )
2
2
2
2
12 3 1
2
14 1
2
+
−
=
12 3
21
42
+
−
= 4
7
2
= 8
7
20. (c)
+ = 90ᵒ , = ⇒ = 90ᵒ, = 30ᵒ
∴ = 90ᵒ - 30ᵒ = 60ᵒ
Now cos2+sin2+tan2+cot2
= cos2ᵒ+sin2ᵒ+tan2ᵒ+cot2ᵒ
= ( )2 22
21 3 13
2 2 3
+ + +
= 1 3 1
34 4 3
+ + + = 1 13
43 3
+ =
21. (a)
2
2
tan 30
1 cot 60
x x−
+ = sin230ᵒ+4cot245ᵒ - cosec230ᵒ
2
2
1
3
11
3
x x
−
+
= ( ) ( )2
2 214 1 2
2
+ −
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⇒ 13 4 4
1 41
3
xx −
= + −
+
⇒
2
134 4
3
x
= ⇒ 1
2 4
x= ⇒ x =
1
2
∴ x2 = 2
1 1
2 4
=
22. (a)
1
1 sin 1 sin
+
+ − =
1
1 sin60 1 sin60
+
+ −
= 1 1
3 31 1
2 2
+
+ −
= 2 2
2 3 2 3+
+ −
= ( )( )4 2 3 4 2 3
2 3 2 3
− + +
+ − =
8
1 = 8
23. (d)
cosecA = 2, ∴ A = 30ᵒ
=
1
1 sin 30 1 21tan 30 1 cos30 3
13 2
+ = +
+ +
= 3 + 1
2 3+ =
2 3 3 1
2 3
+ +
+ =
4 2 3
2 3
+
+ = 2
24. (a)
sin(A+2B) = 3
2 ⇒ (A+2B) = 60ᵒ ............ (i)
and cos (A+4B) ⇒ (A+4B) = 90ᵒ ........... (ii)
(ii)-(i)
2B = 30ᵒ ⇒ B = 15ᵒ
Now, A+2×150 = 60ᵒ ⇒ A = 30ᵒ
25. (c)
+ = 2
, cos =
3
2, = 30ᵒ =
6
Now = 2
-
6
=
3
⇒ ∴ sec = sec
3
= 2
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James Raja
007
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Addition
Partnership Magical
Rex Prime
MatheMagica
Raja
Prime
Partnership
Mantel Maths
Raja
007
Bond
Magical
Rose
James Raja
007
Collection
Addition
Bond Magical
Rex Prime
Partnership
Raja
Prime
Partnership
Mantel Maths
Raja
007
Bond
Magical
Rose
James Raja
007
Collection
Addition
Bond Magical
Rex Prime
Partnership
Raja
Prime
Partnership
Mantel Maths
Raja
007
Bond
Magical
Trigonometry
Conversion T-Rations
Chapter - 04
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34
2
3
2
0, 2𝜋 𝜋
2nd quadrant
sin
cosec (+ve)
1st quadrant
All (+ve)
cos
sec (+ve)
4th quadrant
tan
cot (+ve)
3rd quadrant
♦ Trigonometric ratios change for the odd multiples of 90ᵒ
Ex. 90ᵒ, 270ᵒ, 450ᵒ, 630ᵒ ................ etc
sin ↔ cos
tan↔ cot
sec ↔ cosec
♦ Trigonometric ratios do not change for the even
multiples of 90ᵒ
Ex. 180ᵒ, 360ᵒ, 540ᵒ, 720ᵒ etc
90ᵒ
270ᵒ
0, 360ᵒ
180ᵒ
2nd quadrant
90+
180 -
1st quadrant
90 -
360+
-
270+
360-
4th quadrant
180+
270-
3rd quadrant
1st quadrant
90 - is in 1st quadrant and ratios will change.
In first quadrant all are positive
sin (90 - ) = cos
cos (90 - ) = sin
tan(90 - ) = cot
cot (90 - ) = tan
sec (90 - ) = cosec
cosec (90 - ) = sec
2nd quadrant
( sin, cosec → positive )
for (90+), ratios change
sin (90 + ) = cos
cos (90 + ) = - sin
tan(90+) = - cot
cot (90+ ) = - tan
sec (90 + ) = - cosec
cosec (90 + ) = sec
for (180-), ratios don’t change
sin (180 - ) = sin
cos (180 - ) = -cos
tan(180 - ) = - tan
cot (180 - ) = -cot
sec (180 - ) = - sec
cosec (180 - ) = cosec
3rd quadrant
( tan, cot → positive )
For (180+), ratios don’t change
sin (180+) = - sin
cos (180+ ) = -cos
tan(180+) = tan
cot (180+) = cot
sec (180+) = - sec
cosec (180+) = - cosec
For {(270-)→ odd multiple of 90ᵒ}, ratios change
sin (270 - ) = -cos
cos (270 - ) = -sin
tan(270 - ) = cot
cot (270 - ) = tan
sec (270 - ) = - cosec
cosec (270 - ) = - sec
Conversion of T - Ratios 4
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4th quadrant
(sec, cos → positive)
For (270+) → ratios change
sin (270 + ) = - cos
cos (270 + ) = sin
tan(270 + ) = - cot
cot (270 + ) = -tan
sec (270 + ) = cosec
cosec (27 + ) = - sec
(360-) ≌ (-)
For (-), (360-) ratios don’t change
sin ( - ) = -sin
cos (- ) = cos
tan(- ) = - tan
cot (- ) = - cot
sec (- ) = sec
cosec (- ) = - cosec
Ex. Solve
(i) sec (540ᵒ - ) (ii) tan7
2
+
(iii) cos (11𝜋 - ) (iv) sin 9
2
+
Sol. (i) sec (540-)
[ 540ᵒ = 360ᵒ + 180ᵒ∴ (540ᵒ-) is in 2nd quadrant]
= sec (6×90ᵒ - ) = - sec
(ii) tan 7
2
+
⇒
74
2 2
+ = − +
∴ 7
2
+
is in 4th quadrant] = tan 7
2
+
= - cot
(iii) cos(11𝜋-)
[ 11𝜋 - = (5×2𝜋) + (𝜋-) 2nd quadrant] = - cos
(iv) sin 9
2
+
[
9
2
+ = 2×2𝜋 +
2
+
,
2
+ is in 2nd quadrant]
= cos
Ex. Find value: -
(i) sin 120ᵒ (ii) cos150ᵒ
(iii) tan 225ᵒ (iv) sec3
4
(v) cosec 7
4
(vi) cot
9
4
Sol. (i) sin120ᵒ = sin(90ᵒ+30ᵒ) = cos (30ᵒ)= 3
2
(ii) cos150ᵒ = cos (180-30ᵒ) = -cos(30)= 3
2−
(iii) tan225ᵒ = tan(180ᵒ+45ᵒ) = tan45ᵒ = 1
(iv) sec3
4
= sec
4
−
= - sec
4
= 2−
(v) cosec7
4
= cosec 2
4
−
= - cosce
4
= 2−
(vi) cot9
4
= cot 2
4
+
= cot
4
= 1
Ex. Find value:
(i) sin (750ᵒ) (ii) cos (960ᵒ)
(iii) tan 675ᵒ (iv) cot21
4
(v) sec53
4
Sol. (i) sin(750ᵒ) = sin (2×360ᵒ+30ᵒ) = sin 30ᵒ = 1
2
(ii) cos (960ᵒ) = cos (5×180ᵒ+60ᵒ) = - cos60ᵒ = -1
2
(iii) tan 675ᵒ = tan(2×360ᵒ-45ᵒ) = - tan 45ᵒ = -1
(iv) cot21
4
= cot 5
4
+
= cot
4
= 1
(v) sec53
4
= sec 13
4
+
= - sec
4
= 2−
Ex. If A+B+C = 2
then value of sin (A+B) is
(a) sinC (b) cosC (c) – sinC (d) – cosC
Sol. (b)
A+B+C =2
⇒ A+B =
2
- C
⇒ sin(A+B) = sin2
C
−
= cosC
Trigonometry ratios of complementary angles
If sin x = cos y then x + y = 900
sin x = cos y = sin (900 - y)
x = 900 – y x + y = 900
Thus
If tan x = tan y then x + y = 900
and sec x = sec y then x + y = 900
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Ex. ( )cos37
sin 90 37
− =?
Sol. ( )cos37
sin 90 37
− =
cos37
cos37
= 1
Ex. cosec25ᵒ - sec65ᵒ =?
Sol. cosec25ᵒ - sec65ᵒ = cosec25ᵒ - sec (90-25ᵒ)
= cosec25ᵒ - cosec25ᵒ = 0
Ex. cos80
sin10
+ cos59ᵒcosec31ᵒ =?
Sol. cos80
sin10
+ cos59ᵒcosec31ᵒ
= ( )
cos80
sin 90 80
− +cos59ᵒcosec(90-59ᵒ)
= cos80
cos80
+cos59ᵒsec59ᵒ = 1+
cos59
cos59
= 1+1 = 2
Ex. sin2 21° + sin2 69° is equal to:
Sol. sin2 21° + sin2 69° = sin2 21° + sin2 (900 - 21°)
= sin2 21° + cos2 21° = 1
Ex. sin2 1° + sin2 2° + …… + sin2 89° + sin2 90° = ?
Sol. Let the number of terms be n, then
n = last term first term
common difference
−+ 1
n = 90 1
1
−+ 1 =90
sin21° + sin2 2° + … + sin2 45° + … + sin2 89° + sin2 90°
= (sin2 1° + sin2 89°) + (sin2 2° + … + sin2 88°) + … + to
44 terms + sin2 45°+ sin2 900
= (sin2 1° + cos2 1°) + (sin2 2° + cos220) + … + to 44
terms + sin2 45°+ sin2 900 = 44 + 1
2+ 1
= 2 21
45 sin cos 12
+ =
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1. value of cos(675ᵒ) is
(a) 1
2 (b)
1
2− (c)
1
2 (d)
3
2
2. Value of tan 1650ᵒ is
(a) 3
2 (b)
1
3 (c)
1
2− (d)
1
3
3. cos(-660ᵒ) = ?
(a) 3
2 (b)
1
2 (c)
1
2− (d)
1
2+
4. which of the following is true?
(a) sec 480ᵒ = 2 (b) sin 135ᵒ = 1
2
(c) cos 210ᵒ = 3
2− (d) All of these
5. Which of the following is wrong?
(a) tan (180ᵒ– ) = - tan (b) sin (360 - ) = - sin
(c) sec (270ᵒ - ) = - cosec (d) cos (270+) = - sin
6. cot 3
2
−
= ?
(a) tan (b) - tan (c) cot (d) - cot
7. Value of tan225ᵒcot405ᵒ + tan765ᵒcot675ᵒ is
(a) 1 (b) 0 (c) -1 (d) -2
8. Which of the following is wrong?
(a) sin (-65ᵒ) = -cos25ᵒ
(b) cosec (-225ᵒ) = - cosec45ᵒ
(c) tan (135ᵒ) = - tan45ᵒ
(d) cos 18ᵒ = sin72ᵒ
9. Which of the following is wrong?
(a) tan (𝜋 - ) = - tan (b) cot (-) = - cot
(c) cos (-) = cos (d) sec (-) = - sec
10. The line that makes 640ᵒ angle, lies in
(a) 1st quadrant (b) 2nd quadrant
(c) 3rd quadrant (d) 4th quadrant
11. If cos = 1
2 then value of is:
(a) 60ᵒ, 120ᵒ (b) -60ᵒ, -120ᵒ
(c) 60ᵒ, 300ᵒ (d) 120ᵒ, -300ᵒ
12. If A+B+C = 2
then value of sec (B+C)
(a) cosA (b) secA (c) cosecA (d) - cosecA
13. If A+B+C = 𝜋 then value of tan 2
A B+
is
(a) tan C (b) tan2
C (c) - cot
2
C (d) cot
2
C
14. Which of the following is right
(a) cosec2 (90ᵒ - ) – tan2 = 1
(b) sec2 (90ᵒ - ) – cot2 = 1
(c) sin2 (90ᵒ - ) + sin2 = 1
(d) All of these
15. sinA cos (90ᵒ+A) + cosA sin(270ᵒ+A) = ?
(a) 1 (b) - 1 (c) 0 (d) 2
16. sin 18ᵒ + sin 54ᵒ + sin 150ᵒ + sin 198ᵒ + sin 234ᵒ=?
(a) 1
2− (b)
3
2 (c)
1
2 (d) -
3
2
17. sin + cos = 3 sin (90ᵒ-) then cot = ?
(a) 3 1
2
+ (b)
3 1
2
− (c) 3 (d)
3
2
18. ( ) ( )
( )
sin 90 cos 90
cot 90
− −
− =?
(a) 1+sin2 (b) 1- sin2 (c) – sin2 (d) 2- sin3
19. ( ) ( ) ( )
( )
cos 90 sin sin 90 cos sin 90 sin
cos cos 90
− − −+
−
(a) 1 (b) 2 (c) -1 (d) 3
20. Evaluate tan54 cos 32 sin 41
cot 36 sec58 cos49
ec + +
Exercise - 1
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(a) 1 (b) 0 (c) 3 (d) 3
2
21. sin 33 cos57
cos57 sin 33
+
- 4 sin245ᵒ
(a) 1 (b) 1
2 (c) -1 (d) 0
22. cos9ᵒ + cot81ᵒ = ?
(a) cos81ᵒ + tan9ᵒ (b) sin81ᵒ + tan9ᵒ
(c) sin81ᵒ + cos81ᵒ (d) sin9ᵒ+tan9ᵒ
23. Value of tan14ᵒ tan43ᵒ tan 47ᵒ tan76ᵒ is
(a) 2 (b) 3 (c) 1 (d) 4
24. The value of cot10ᵒ cot20ᵒ cot60ᵒ cot70ᵒ cot80ᵒis
(a) 1 (b) -1 (c) 3 (d) 1
3
25. if sin sec29ᵒ = 1, then value of is
(a) 29ᵒ (b) 61ᵒ (c) 31ᵒ (d) 59ᵒ
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1. (sin 670 + cos 230) (sin 670 - cos 230)
(a) 3 (b) 0 (c) 1 (d) 2
2. If tan (7θ + 280) = cot (300 - 3θ) then value of θ is
(a) 80 (b) 50 (c) 600 (d) 90
3. If x, y are positive acute angles and x + y < 900 and sin
(2x - 200) = cos (2y + 200), then value of sec (x + y) is
(a) 2 (b) 1
2 (c) 1 (d) 0
4. sin (x + y) = cos [3(x + y)] then value of cot2(x + y) is
(a) 3 (b) 1 (c) 0 (d) 1
3
5. If (2θ + 450) and 3θ are acute angles and tan (2θ + 450)
= cot3θ, then the value of θ is:
(a) 50 (b) 90 (c) 120 (d) 150
6. If sec (4θ - 500) = cosec (500 - θ) and 00 < θ < 900 then
the value of θ is
(a) 3301
3 (b) 180 (c) 23
01
3 (d) 300
7. If tan 7θ tan 2θ = 1, then the value of tan 3θ is
(a) 3 (b) 1
3− (c)
1
3 (d) 3−
8. Value of (tan 10 tan 20 tan 30 ..... tan 890)
(a) 1 (b) 245 (c) Not defined (d) 0
9. The value of cos 10 cos 20 cos 30..... cos 1770 cos 1780 cos
1790
(a) 0 (b) 1
2 (c) 1 (d)
1
2
10. The value of cot 420 cot 430 cot 440 cot 450 cot 460 cot
470 cot 480 is
(a) 0 (b) 1 (c) - 1 (d) 3
11. ( )
( ) ( )
0 0 0
0 0 0
sin300 tan240 sec 390
cos210 cot 135 cosec 315
− − −
=?
(a) 3 (b) 2 (c) 2
3 (d)
3
2
12. If sin ∝ sec (300 + ∝) = 1 (00< ∝ < 600) then sin ∝ + cos
2∝ =?
(a) 1 (b) 2 3
2 3
+ (c) 0 (d) 2
13. If A, B and C be the angles of a triangle, then which of
the following is incorrect?
(a) sin A B
2
+
= cosC
2 (b) cos
A B
2
+
= sinC
2
(c) tan A B
2
+
= sec C
2 (d) cot
A B
2
+
= tanC
2
14. If Q is a positive acute angle and tan2θtan3θ = 1 then
the value of (2 cos25
2
-1) is
(a) -1
2 (b) 1 (c) 0 (d)
1
2
15. If A + B = 900, then
+
−2
2
tanAtanB tanAcotB sin B
sinAsecB cos A
(a) tan A (b) cot B (c) tan B (d) sin B
16. If tan150 = 2 3− then tan 150 cot 750 + tan 750 cot 150
=?
(a) 14 (b) 12 (c) 10 (d) 8
17. Find value of cot20
cot
3
20
cot
5
20
cot
7
20
cot
9
20
(a) - 1 (b) 1
2 (c) 0 (d) 1
18. tan (x + y) tan (x - y) = 1 then tan 2x
3
= ?
(a) 1
3 (b)
2
3 (c) 3 (d) 1
19. If A = tan110 tan290, B = 2 cot610cot790 then which of
Exercise - 2
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40
the following is true?
(a) A = 2B (b) A = - 2B (c) 2A = B (d) 2A = - B
20. sin2 5° + sin2 6° + …… + sin2 84° + sin2 85° = ?
(a) 1
392
(b) 1
402
(c) 40 (d) 1
392
21. The value of sin2 5° + sin2 10° + sin2 15° + …… + sin2 85°
+ sin2 90° is:
(a) 1
72
(b) 1
82
(c) 1
102
(d) 1
102
22. The value of sin2 25° + sin2 45° + sin2 65° + sin2 85° is
equal to:
(a) 1.5 (b) 2 (c) 2.5 (d) 3
23. The value of sin2 1° + sin2 5° + sin2 9° + …… + sin2 89°:
(a) 1
112
(b) 11 2 (c) 11 (d) 11
2
24. If sin2 θ - 2cos θ + 1
4= 0 then θ =? (00 < θ < 900)
(a) 3
(b)
6
(c)
2
(d)
3
2
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Answer key
1 a 2 d 3 b 4 d 5 d
6 b 7 b 8 b 9 d 10 d
11 c 12 c 13 d 14 d 15 b
16 c 17 a 18 b 19 a 20 c
21 b 22 d 23 c 24 d 25 b
1. (a)
cos (675ᵒ) = cos (2×360ᵒ-45ᵒ)
[675ᵒ= 720ᵒ-45ᵒ ∴ 4th quadrant] = cos45ᵒ = 1
2
2. (d)
tan(1650ᵒ)
[here 1650ᵒ = (360×4+210ᵒ)
∴ 1650ᵒ is in 3rd quadrant and in 3rd quadrant tan and
cot are positive]
= tan (360ᵒ×4 + 210ᵒ) = tan(210ᵒ) = tan(180ᵒ+30ᵒ)
= tan30ᵒ =1
3
3. (b)
cos(-) = cos
∴ cos(-660ᵒ) = cos660
cos(660ᵒ) = cos(2×360ᵒ - 60ᵒ) [4th quadrant]
= cos60ᵒ = 1
2
4. (d)
Option (a)
sec480ᵒ = sec(360ᵒ+120ᵒ) = sec(5×90ᵒ+30ᵒ)
= - cosec30ᵒ (2nd quadrant) = - 2 (true)
Option (b)
sin135ᵒ = sin (90+45ᵒ)
= cos45ᵒ (2nd quadrant) = 1
2 (true)
Option (c)
cos210ᵒ = cos(180ᵒ+30ᵒ)
= - cos30ᵒ (3rd quadrant) = 3
2− (true)
5. (d)
cos (270+) = sin
(270ᵒ+) is in 4th quadrant and cos is positive in 4th
quadrant
270ᵒ is odd multiple of 90ᵒ
∴ cos → sin
6. (b)
cot 3
2
−
= cot -
3
2
−
= - cot 3
2
−
[ as cot (-) = - cot ] = - tan
7. (b)
tan225ᵒ cot405ᵒ + tan 765ᵒcot675ᵒ
= tan (180ᵒ+ 45ᵒ) cot (360ᵒ+ 45ᵒ) + tan (720ᵒ+ 45ᵒ) cot
(720ᵒ- 45ᵒ) = tan45ᵒ cot45ᵒ + tan45ᵒ (-cot45ᵒ)
= tan45ᵒ cot45ᵒ - tan45ᵒ cot 45ᵒ = 0
8. (b)
Option (a)
sin (-65ᵒ) = - sin65ᵒ = - sin(90ᵒ-25ᵒ) = - cos25ᵒ
Option (b)
cosec (-225ᵒ) = - cosec 225ᵒ
= - cosec (180ᵒ+45ᵒ) = cosec45ᵒ
Option (c) tan135ᵒ = tan (180ᵒ- 45ᵒ) = - tan45ᵒ
Option (d)
cos18ᵒ = cos (90ᵒ-72ᵒ) = sin72ᵒ
9. (d)
sec and cos are positive in 4th quadrant hence sec (-) =
sec and cos(-) = cos
10. (d)
4th quadrant
Solutions
Exercise - 1
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640ᵒ = 360ᵒ×2 - 80ᵒ
∴ 4th quadrant
11. (c)
cos = 1
2 = cos(60ᵒ) = cos(360ᵒ-60ᵒ) = cos (300ᵒ)
∴ = 60ᵒ, 300ᵒ
12. (c)
A+B+C = 2
⇒ B+C =
2
-A
⇒ sec (B+C) = sec 2
A
−
⇒ sec (B+C) = cosecA
13. (d)
A+B+C = 𝜋
A+B = 𝜋 – C ⇒ 2
A B+=
2 2
C−
⇒ tan2
A B+
= tan2 2
C −
= cot
2
C
14. (d)
Option (a)
cosec2 (90ᵒ - ) – tan2
sec2 - tan2 = 1
Option (b)
sec2(90ᵒ - ) – cot2
cosec2 - cot2 = 1
Option (c)
sin2(90ᵒ - ) + sin2 = 1
cos2 + sin2 = 1
15. (b)
sin A cos(90ᵒ+A) + cosA sin(270ᵒ+A)
= sin A (-sinA) + cos (-cosA)
= -[sin2A + cos2A] = - 1
16. (c)
sin18ᵒ + sin54ᵒ + sin150ᵒ + sin198ᵒ + sin 234ᵒ
= sin18ᵒ + sin54ᵒ + sin (180ᵒ-30ᵒ) + sin(180ᵒ+18ᵒ) + sin
(180ᵒ+54ᵒ)
= sin18ᵒ + sin54ᵒ + sin30ᵒ + (-sin18ᵒ) + (-sin54ᵒ)
= sin30ᵒ = 1
2
17. (a)
sin + cos = 3 sin (90ᵒ-)
⇒ sin + cos = 3 cos
⇒ sin = ( )3 1− cos ⇒ cos 1
sin 3 1
=
−
cot = ( )
( )( )
3 11 3 1
23 1 3 1
+ +=
− +
18. (b)
( ) ( )
( )
sin 90 cos 90
cot 90
− −
−
= cos sin
tan
=
cos sin
sin
cos
=cos2 = 1-sin2
19. (a)
( ) ( ) ( )( )
cos 90 sin sin 90 cos sin 90 sin
cos cos 90
− − −+
−
= sin sin cos cos cos sin
cos sin
+
= sin2 + cos2 =1
20. (c)
tan54 cos 32 sin 41
cot 36 sec58 cos49
ec + +
= ( ) ( ) ( )
tan54 cos 32 sin 41
cot 90 54 sec 90 32 cos 90 41
ec + +
− − −
=tan54 cos 32 sin 41
tan54 cos 32 sin 41
ec
ec
+ +
= 3
21. (d)
sin 33 cos57
cos57 sin 33
+
- 4 sin245ᵒ
sin33ᵒ = sin(90ᵒ-57ᵒ) = cos57ᵒ
∴
2
cos57 cos57 14
cos57 cos57 2
+ −
= 1 + 1 - 2 = 0
22. (b)
cos9ᵒ + cot81ᵒ
= cos(90ᵒ - 81ᵒ) + cot(90ᵒ - 9ᵒ) = sin81ᵒ + tan9ᵒ
23. (c)
tan14ᵒ tan43ᵒ tan 47ᵒ tan76ᵒ
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= tan14ᵒ tan43ᵒ tan (90ᵒ-43ᵒ) tan(90ᵒ-14ᵒ)
= tan14ᵒ tan43ᵒ cot43ᵒ cot14ᵒ
= (tan14ᵒcot14ᵒ)(tan43ᵒcot43ᵒ) = 1
1
tancot
=
24. (d)
cot10ᵒ cot20ᵒ cot60ᵒ cot70ᵒ cot80ᵒ
= cot10ᵒ cot20ᵒ cot60ᵒ cot(90ᵒ-20ᵒ) cot(90ᵒ-10ᵒ)
= cot10ᵒ cot20ᵒ cot60ᵒ tan20ᵒ tan10ᵒ
= (cot10ᵒtan10ᵒ) (cot20ᵒtan10ᵒ) cot60ᵒ
= 1×1×1
3 =
1
3
25. (b)
sin sec29ᵒ = 1 ⇒ sinᵒ = 1
sec 29 = cos29ᵒ
⇒ sin = cos (90ᵒ- 61ᵒ) = sin61ᵒ
∴ = 61ᵒ
Answer key 1 b 2 a 3 a 4 b 5 b
6 d 7 c 8 a 9 a 10 b
11 b 12 a 13 c 14 c 15 a
16 a 17 d 18 a 19 c 20 b
21 d 22 c 23 a 24 a
1. (b)
(sin 670 + cos 230) (sin 670 - cos 230)
= [sin 670 + cos (90 - 670)] [sin670 - cos(90 - 670)]
= (sin 670 + sin 670) (sin 670 - sin 670) = 2sin670 × 0 = 0
2. (a)
tan (7θ + 280) = cot (300 - 3θ)
⇒ tan (7θ + 280) = cot [900 - (600 + 3θ)]
⇒ tan (7θ + 280) = tan (600 + 3θ)
⇒ 7θ + 280 = 600 + 3θ ⇒ 4θ = 320 ⇒ θ = 80
[If tan x = cot y then x + y = 900]
3. (a)
sin (2x - 200) = cos (2y + 200)
⇒ sin (2x - 200) = cos [900 - (700 - 2y)]
⇒ sin (2x - 200) = sin (700 - 2y)
⇒ 2x - 200 = 700 - 2y
⇒ 2x + 2y = 900 ⇒ x + y = 450
sec (x + y) = sec 450 = 2
(sin A = cos B, then A + B = 900)
here 2x - 200 + 2y + 200 = 900 ⇒ x + y = 450
4. (b)
sin (x + y) = cos [3 (x + y)]
= sin [900 - 3(x + y)] ⇒ x + y = 900 - 3(x + y)
⇒ 4(x + y) = 900 ⇒ 2(x + y) = 450
Now cot2(x + y) = cot 450 = 1
5. (b)
tan (2θ + 450) = cot 3θ
∴ (2θ + 450) + 3θ = 900 ⇒ 5θ = 450 ⇒ θ = 90
6. (d)
sec (4θ - 500) = cosec (500 - θ)
⇒ sec (4θ - 500) = cosec [90 - (400 + θ)]
⇒ sec (4θ - 500) = sec (400 + θ)
⇒ 4θ - 500 = 40 + θ ⇒ 3θ = 900 ⇒ θ = 300
(sec A = cosec B 00 < A, B < 900
then A + B = 900)
Exercise - 2
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here 4θ - 500 + 500 - θ = 900 ⇒ 3θ = 900 ⇒ θ = 300
7. (c)
tan7θ tan 2θ = 1
⇒ tan7θ = 1
tan2= cot 2θ
⇒ tan 7θ = tan (90 - 2θ) ⇒ 7θ = 900 - 2θ
⇒ 9θ = 900 ⇒ θ = 100
Now tan 300 = 1
3
8. (a)
tan 10 tan 20 tan 30..... tan 890
= (tan10 tan890)(tan20 tan880)..... tan450
= (tan10cot10)(tan20 cot20) ....... tan 450
= 1 × 1 × ......... × 1 = 1
9. (a)
cos 10 cos 20 cos 30 ..... cos 1770 cos 1780 cos 1790
= cos 10 cos 20 cos 30..... cos 900 ....... cos 1780 cos 1790
= 0 (∵cos900 = 0)
10. (b)
cot 420 cot 430 cot 440 cot 450 cot 460 cot 470 cot 480
= cot 420 cot 430 cot 440 cot 450 tan 420 tan 430 tan 440
= (cot 420tan 420)(cot 430tan 430)(cot 440tan 440 )cot
450 = 1
11. (b)
sec (-θ) = sec θ
= ( ) ( ) ( )( )( )( )
− +
+ − −
0 0 0 0 0
0 0 0 0
sin 360 60 tan 180 60 sec 390
cos 180 30 cot135 cosec315
= ( ) ( )
( ) ( )
0 0 0 0
0 0 0 0 0
sin60 tan60 sec 360 30
cos30 cot 180 45 cosec 360 45
− +
− − −
= ( )( )
0 0 0
0 0 0
sin60 tan60 sec30
cos30 cot 45 cosec45
−
− − −=
3 23
2 3
31 2
2
= 2
12. (a)
sin ∝ = ( )0
1
sec 30 + ⇒ sin ∝ = cos (300 + ∝)
∴ ∝ + 300 + ∝ = 900 ⇒ 2∝ = 600, ∝ = 300
Now sin ∝ + cos 2∝ = sin300 + cos 600
= 1 1
2 2+ = 1
13. (c)
A + B + C = π
then A B C
2 2 2
+ = −
tan A B
2
+
= tan C
2 2
−
⇒ tan
A B
2
+
= cot C
2
14. (c)
tan 2θ tan 3θ = 1
tan2θ = cot3θ
∴ 2θ + 3θ = 900
5θ = 900
θ = 180
Now
02 5 18
2 cos 12
−
= 2 cos2450 - 1 = 2 ×
21
2
- 1 = 0
15. (a)
+−
2
2
tanAtanB tanAcotB sin B
sinAsecB cos A
= ( ) ( )( )
( )0 0 2 0
20
tanAtan 90 A tanAcot 90 A sin 90 A
cos AsinAsec 90 A
− + − −−
−
= 2 2
2
tanAcot A tan A cos A
sinAcosecA cos A
+−
= 21 tan A 1+ − = 2tan A = tanA
16. (a)
tan150 = 2 3− = tan (900-750) = cot750
cot150 = ( )( )0
1 1 2 3
tan15 2 3 2 3 2 3
+= =
− − +
= 2 3
4 3
+
−= 2 3+
cot 150 = tan 750
now tan 150 cot 750 + tan 750 cot 150
= ( )2 3− ( )2 3− + ( )2 3+ ( )2 3+
= 4 + 3 - 2 3 + 4 + 3 + 2 3 = 14
17. (d)
cot20
cot
3
20
cot
5
20
cot
7
20
cot
9
20
= cot 20
cot
3
20
cot
4
cot
3
2 20
−
cot
2 20
−
= cot20
cot
3
20
cot
4
tan
3
20
tan
20
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=
3tan cot cot tan cot
20 20 20 20 4 = 1
18. (a)
tan (x + y) tan (x - y) = 1
tan (x + y) = ( )1
tan x y− ⇒ tan (x + y) = cot (x - y)
∴ (x + y) + (x - y) = 900 ⇒ 2x = 900,
now tan
090
3
= tan 300 =1
3
19. (c)
A = tan110tan290
B = 2cot610cot790 = 2 cot(900 - 290) cot (900-110)
= 2 tan290 tan110 = 2A
∴ B = 2A
20. (b)
Let the number of terms be n, then
n = last term first term
common difference
−+ 1
n = 85 5
1
−+ 1 = 81
sin2 5° + sin2 6° + … + sin2 45° + … + sin2 84° + sin2 85°
= (sin2 5° + sin2 85°) + (sin2 6° + … + sin2 84°) + … + to
40 terms + sin2 45°
= (sin2 5° + cos2 5°) + (sin2 6° + … + sin2 45°) + … + to 40
terms + sin2 45°
= 40 + 2 21 1
40 sin cos 12 2
= + =
21. (d)
no. of terms = last term first term
common difference
− + 1
= 90 5
5
− + 1 = 17 + 1 = 18
(sin2 45° + sin2 85°) + (sin2 10° + sin2 80°) + … to 8
terms + sin2 45° + sin2 90°
= (sin2 5° + cos2 5°) + (sin2 10° + cos2 10°) + … to 8
terms + 1
2 + 1 = 8 +
1
2 + 1 =
19
2
[ sin (90° – ) = cos ; sin2 + cos2 = 1]
22. (c)
sin2 25° + sin2 85° + sin2 25° + sin2 65° + sin2 45°
= sin2 5° + sin2 (90° – 5°) + sin2 25° + sin2 (90° – 25°) +
sin2 45°
= sin2 5° + cos2 5° + sin2 25° + cos2 25° = sin2 45°
= 1 + 1 + 1 1
2 2.52 2
= =
23. (a)
n = last term first term
common difference
−+ 1 ⇒ n =
89 1
4
−+ 1 = 23
sin2 1° + sin2 89° + sin2 5° + sin2 85° + … + to 22 terms +
sin2 45°
= (sin2 1° + cos2 1°) + (sin2 5° + cos2 5°) + … + to 11
terms +
2
1
2
= 2 21 1
11 11 sin cos2 2
+ = +
24. (a)
sin2 θ - 2cos θ + 1
4= 0
⇒ 1-cos2 θ - 2 cos θ + 1
4= 0 ⇒ cos2 θ + 2 cos θ -
5
4= 0
cos θ =
− + 5
2 4 4 14
2
= 2 9
2
− =
2 3 1
2 2
− = (0 < θ < 900)
∴ θ = 3
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* * * Some Basic Identities
h
b
P
h2 = p2 + b2 .......... (i)
eq. (i) divide both sides by h2
1 = 2 2
2 2
p b
h h+
⇒ sin2 + cos2 = 1
eq.(i) divide both side by b2
2 2
h p
b b
=
+ 1
sec2 = tan2+1
⇒ sec2 - tan2 = 1
eq. (i) divide both sides by p2
2 2
1h b
p p
= +
⇒ cosec2 = 1+ cot2
⇒ cosec2 − cot2 = 1
As sec2 - tan2 = 1
From a2 – b2 = (a-b)(a+b)
(sec - tan) (sec + tan) = 1
⇒ sec + tan = 1
sec tan −
and
cosec2 - cot2 = 1
⇒ (cosec - cot) (cosec + cot) = 1
⇒ cosec + cot = 1
cos cotece −
***
If a cos - bsin = c,
and asin + bcos = k
then a2+b2 = c2+k2
Proof: a cos - b sin = c ..... (i)
a sin + b cos = k ..... (ii)
(i)2 + (ii)2
a2cos2 + b2sin2 - 2ab sin cos + a2sin2 + b2cos2 +
2ab sin cos = c2+k2
a2 (cos2 + sin2 ) + b2 (cos2 + sin2 ) = c2 + k2
⇒ a2 + b2 = c2 k2 ⇒ k = 2 2 2a b c + −
Basic Identities 5
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Ex.1 sec2 θ + cosec2 θ =?
Sol. sec2 θ + cosec2 θ
= 2 2
2 2 2 2
1 1 sin cos
cos sin cos sin
+ + =
= 2 2 2 2
1 1 1
cos sin cos sin=
= sin2 θ cosec2 θ
Ex.2 (1 + cot2θ) (1 – cos θ) (1 + cos θ) =?
Sol. (1 + cot2 θ) (1 - cos θ) (1 + cos θ)
= cosec2 θ (1 - cos2 θ) = cosec2 θ sin2 θ = 1
Ex.3 cos2 θ cosec θ + sin θ =?
Sol. cos2 θ cosec θ + sin θ
= cos2 θ × 1
sin + sin θ =
2 2cos sin 1
sin sin
+ =
= cosec θ
Ex.4 (1 + tan2 A) cos2 A =?
Sol. (1 + tan2 A) cos2 A
= sec2 A cos2 A [∵ 1 + tan2 A = sec2 A]
= 2
1
cos A× cos2 A = 1
Ex.5 ( cosec2 θ - 1) tan2 θ =?
Sol. (cosec2 θ - 1) tan2 θ
= cot2 θ × tan2 θ [∵ cosec2 θ = 1 + cot2 θ]
= 1 [∵ tan θ × cot θ = 1]
Ex.6 sinA cosA
cosecA secA+ =?
Sol. sinA cosA
cosecA secA+
= sin A × 1
cosecA+ cos A ×
1
secA
= sin A × sin A + cos A × cos A
= sin2 A + cos2 A = 1
Ex.7 1 1
1 cosA 1 cosA+
− + =?
Sol. 1 1
1 cosA 1 cosA+
− +=
( )( )1 cosA 1 cosA
1 cosA 1 cosA
+ + + −
− +
= 2 2
2 2
1 cos A sin A=
− = 2 cosec2 A
Ex.8 sec–tan = 4 then find sec =
Sol. sec - tan = 4 .... (i)
The sec + tan = 1
4 ...... (ii)
Eq.(i) + eq. (ii)
2sec = 4+1
4 ⇒ sec =
17
8
Ex.9 cosec + cot = 3 then find cos =?
Sol. cosec + cot = 3 ...... (i)
Then cosec - cot =1
3 .......... (ii)
(i) + (ii)
2cosec = 3+1
3=
10
3 ⇒ cosec =
5
3
5
4
3
cos = 4
5
b
h=
Examples
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1. (1 + tan2θ) (1 + sin θ) (1 – sin θ)
(a) 0 (b) 1 (c) 2 cosec2θ (d) 2tan2θ sin2θ
2 2
2
1 tan
cosec
+
=?
(a) cot2θ (b) sin2θ (c) cosec2θ (d) tan2θ
3. 2 2
1 1
1 tan 1 cot+
+ +
(a)cosec2θ (b) 0 (c) 2sec2θ (d) 1
4. 2 2
2 2
tan sec
cot cosec
−
− =?
(a) 1 (b) - 4 (c) 4 (d) 7
5. tan2 θ - sin2 θ = ?
(a) tan θ sin θ (b) cot2 θ - cos2 θ
(c) sin2 θ tan2 θ (d) sec2 θ + cos2 θ
6. sin2 A cos2 B - cos2 A sin2 B = ?
(a) sin2A + cos2B (b) sin2A - sin2B
(c) cos2A - sin2B (d) sin2A - cos2B
7. ( )2
2
1 cot tan
sec
+
=?
(a) cos θ (b) cot θ (c) cosec2 θ (d) tan2 θ
8. (cosec2 θ - 1)(sec2 θ - 1) =?
(a) 1 (b) 2 (c) 4 (d) 3
9. cos2 θ + cos2 θ cot2 θ =?
(a) tan2 θ (b) cot2 θ (c) cot2θ (d) tan2θ
10. (cosec θ - sin θ)(sec θ - cos θ)(tan θ + cot θ) = ?
(a) 2 (b) sec2 θ (c) tan2 θ (d) 1
11. sec 1
sec 1
−
+=?
(a) 2
cos
1 cos
+ (b)
2cos
1 sin
−
(c) 2
sin
1 cos
+ (d)
−
21 cos
sin
12. tan θ + cot θ =?
(a) sinθ cos θ (b) 2secθ
(c) secθ cosec θ (d) sec2θ
13. + −
+− +
1 cos 1 cos
1 cos 1 cos
(a) 2sec θ (b) 2cot θ
(c) 2cosec θ (d) cosec θ + cot θ
14. tan2A + cot2A = ?
(a) sec2A + cot2A (b) tan2A + cosec2A - 2
(c) sec2A cosec2A - 1 (d) sec2A cosec2A - 2
15. 1 1
sec 1 sec 1−
− +=?
(a) cot2θ (b) 2sec2 θ (c) 2
2
cot (d)
2
2
tan
16. cot2A cosec2B - cot2B cosec2A = ?
(a) cot2A - cot2B (b) cosec2A - cot2B
(c) tan2A + sin2B (d) cot2A + cot2B
17. tan sin
tan sin
+
− =?
(a) sec 1
sec 1
+
− (b)
cosec 1
sec
+
(c) sec
cosec 1
− (d) 1
18. 1 sin
1 sin
−
+ =?
(a) (sec θ + tan θ)2 (b) (sec θ - tan θ)2
(c) (cosec θ - cot θ)2 (d) (cosec θ - cot θ)2
19. 2 2
1 1
cos sin+
=?
(a) sin2 θ cos2 θ (b) 2cosec2 θ
(c) tan2 θ sin2 θ (d) sec2 θ cosec2 θ
Exercise - 1
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20. sec4 θ - sec2 θ =?
(a) tan4 θ - tan2 θ (b) tan2 θ - tan4 θ
(c) tan4 θ + tan2 θ (d) None of these
21. cos4 A + sin2 A cos2 A =?
(a) sec2A (b) cos2A (c) sin2A (d) cos2A
22. sinA cosA sinA cosA
sinA cosA sinA cosA
+ −+
− +
(a) 2
2
2sin A 1− (b)
2
2
2cos A 1−
(c) 0 (d) 1
23. 2 2sec cosec+ =
(a) tan θ + cot θ (b) sin θ + cos θ
(c) cos θ + cot θ (d) tan θ + cos θ
24. sin6 A + cos6 A
(a) 1 + 3sin2A cos2A (b) 1 - 3sin2A + 4cos2A
(c) 1 + 3sin2A - 4cos2A (d) 1 - 3sin2A cos2A
25. cot A tanB
cot B tan A
+
+=?
(a) cot A + tan B (b) cot A tan B
(c) cos B – tan A (d) cot B + tan B
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1. If tan + cot = 2 then value of tan5+cot5 is
(a) 2 (b) 1 (c) 10 (d) 10
2. If sin + cosec = 2 then the value of sinn + cosecn is
(a) 2n (b) 2n (c) 2 (d) 2n+1
3. If cos + sec = 2, then the value of cos9 + sec9 is
(a) 2 (b) 29 (c) 1 (d) 2
4. If tan + cot = 2 then the value of tan-10 + cot-15 =?
(a) -2 (b) 2 (c) 210 (d) 2-5
5. If sin2+sin2 = 2 then cos 2
+
=?
(a) 1 (b) -1 (c) 0 (d) 0.5
6. if sec + tan = 2 then sec = ?
(a) 5
2 (b)
5
4 (c)
7
4 (d)
7
2
7. If cot A + cosec A = 3 and A is acute angle then value of
cot A is
(a) 4
5 (b) 1 (c)
1
2 (d)
3
4
8. If sec + tan = 2+ 5 then value of sin + cos
(a) 5 (b) 7
5 (c)
1
5 (d)
3
5
9. If cosec = x+1
4x (0ᵒ<<90ᵒ) then value of cosec + cot
is
(a) 2
x (b) 2x (c) x (d)
1
4x
10. If 3sin+5cos = 5 then 5 sin - 3cos =?
(a) ±3 (b) ±5 (c) 1 (d) ±2
11. sin - cos = 7
13and 0ᵒ < < 90ᵒ then sin + cos
equals
(a) 17
13 (b)
13
17 (c)
1
13 (d)
1
17
12. If sec2 + tan2 = 7 then the value of is
(a) 60ᵒ (b) 30ᵒ (c) 0ᵒ (d) 90ᵒ
13. If be an acute angle and 7sin2 + 3cos2 = then the
value of tan is
(a) 3 (b) 1
3 (c) 1 (d) 0
14. Find value of (cos+sin)2+ (cos-sin)2
(a) 2 (b) 3 (c) 4 (d) 5
15. If sec2 + tan2 = 7
12then sec4 - tan4 =?
(a) 49
144 (b)
12
7 (c)
7
12 (d) 1
16. sec4 - sec2 = ?
(a) tan2 + cot2 (b) tan4 + cot2
(c) tan4 + tan2 (d) sec2 + cosec2
17. If x = a (sin + cos), y = b (sin - cos) then the value of 2 2
2 2
x y
a b+ is
(a) 0 (b) 1 (c) 2 (d) -2
18. If x = sinA cosB, y = sinAsinB, & z = cosA then
(a) x2 + y2 + z2 = 2 (b) x2 - y2 + z2 = 2
(c) x2 + y2 - z2 = 2 (d) -x2 + y2 + z2 = 2
19. If cos + cos2 = 1 then sin8 + 2sin6+sin4 =?
(a) 0 (b) -1 (c) 1 (d) 2
20. if cot + tan = x and sec - cos = y, then
( ) ( )2 3
2 23 2x y xy− = ?
(a) 4 (b) 3 (c) 2 (d) 1
21. if tan - tan2 = 1 then the value of sec2 - sec4 is
(a) 1 (b) -1 (c) 2 (d) 0
Exercise - 2
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22. the value of 1 1
cos cot sinec −
− is
(a) 1 (b) cot (c) cosec (d) tan
23. cosec - cot = 7
2, then cot = ?
(a) 47
28 (b)
83
28 (c)
49
28− (d)
45
28−
24. If cos2 - sin2 = 1
3 then find the value of (cos4 -
sin4+1)?
(a) 1 (b) 1
3 (c)
4
3 (d)
5
3
25. Value of cot180 (cot720 cos2220 +0 2 0
1
tan72 sec 68) is
(a) 1 (b) 2 (c) 3 (d) 1
3
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1. tan sec 1
tan sec 1
+ −
− +=?
(a) 1 sin
cos
+
(b)
1 cos
sin
+
(c)
1 tan
cos
+
(d) 1
2. cot A cosecA 1
cot A cosecA 1
+ −
− +
(a) 1 sinA
cosA
− (b)
1 cosA
sinA
+ (c)
1 cotA
cosA
− (d) 1
3. sec A (1 – sin A) (sec A + tan A)
(a) - 1 (b) 1 (c) 2 (d) - 3
4. 2cosA sin A
1 tan A sin A cosA+
− −
(a) sinA - cosA (b) sin2A - cos2A
(c) sinA + cosA (d)3sinA + 2cosA
5. 2 3cos sin
1 tan sin cos
+
− −
(a) 1+ sin θ + cos θ (b) 1-sin θ cos θ
(c) 1+sin θ cos θ (d) 1 + sin θ - cos θ
6. 3 3sin cos
sin cos
+
+ + sin θ cos θ
(a) 0 (b) 1 (c) 2 (d) 3
7. ( sin A + sec A)2 + (cos A + cosec A)2 = ?
(a) (1 – sec A cosec A)2 (b) (1 + sec A cosec A)2
(c) (1 + sin A + cos A)2 (d) (1 + sec A - cosec A)2
8. (1 + tan A tan B)2 + (tan A - tan B)2=?
(a) sec2A sec2B (b) sec2A + sec2B
(c) sec2 A - sec2 B (d) sec4 A sec2 B
9. sin θ (1 + tan θ) + cos θ (1 + cot θ) =?
(a) cosec θ - sec θ (b) sec θ - cosec θ
(c) cosec θ + sec θ (d) 1
10. tan6 θ + 3 tan2 θ sec2θ + 1 =?
(a) sec6 θ (b) sin6 θ (c) sec8 θ (d) cosec6 θ
11. 2(sin6 θ + cos6 θ) – 3(sin4 θ + cos4 θ) + 1
(a) 0 (b) 1 (c) - 2 (d) 3
12. tan cot
sin cos
−
=?
(a) tan2 θ + cot2 θ (b) tan2 θ - cot2 θ
(c) cos2 θ - cosec2 θ (d) sin2 θ + cot2 θ
13. sec4A (1-sin4A) - 2 tan2A=?
(a) 2 (b) 3 (c) 5 (d) 1
14. secA tanA
secA tanA
−
+=?
(a) 1 - 2 sec A tanA + 2 tan2 A
(b) 1 + 2 sec A tanA + 2 tan2 A
(c) 1 + 2 sec A tanA - 2 tan2 A
(d) 1 - 2 sec A tanA - 2 tan2 A
15. 2 sec2 θ - sec4 θ - 2 cosec2 θ + cosec4 θ
(a) cot4 θ - tan4 θ
(b) cot4 θ + tan4 θ
(c) cosec4 θ - tan4 θ
(d) cos4 θ + cosec4 θ
16. − +
++ −
1 sin 1 sin
1 sin 1 sin
(a) 2 tan θ (b) 2 cosec θ
(c) sec θ - tan θ (d) 2sec θ
17. 2 cosec2230 cot2670 - sin2230 - sin2670 - cot2670 =?
(a) 1 (b) sec2 230 (c) tan2 230 (d) 0
18. sin cos
1 cot 1 tan
+
− − =?
(a) 0 (b) 1
(c) sin θ + cos θ (d) sin θ - cos θ
19. Value of cos cos
1 sin 1 sin
+
− + is
(a) 2 sec θ (b) sec2 θ (c) 2 cosec θ (d) 2 sin θ
20. sinA 1 cosA
1 cosA sinA
++
+=
(a) 2sinA (b) 2 cosecA (c) tan2
A (d) cot
2
A
Exercise - 3
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21. (1 + cot θ - cosec θ)(1 + tan θ + sec θ) =?
(a) 3 (b) 1 (c) 0 (d) 2
22. tan tan
sec 1 sec 1
−
− +=?
(a) 2sin (b) 2sec
(c) 2 cosec (d) cosec + sec
23. tan2 𝜙 + cot2 𝜙 + 2 =?
(a) 2 (b) sec2 𝜙 sin2 𝜙
(c) cos2 𝜙 cosec2 𝜙 (d) sec2 𝜙 cosec2 𝜙
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Answer key 1 b 2 d 3 d 4 a 5 c
6 b 7 b 8 a 9 b 10 d
11 d 12 c 13 c 14 d 15 d
16 a 17 a 18 b 19 d 20 c
21 d 22 a 23 a 24 d 25 b
1. (b)
(1 + tan2 θ) (1 + sin θ) (1 - sin θ)
= (1 + tan2 θ) (1 - sin2 θ)
= sec2 θ cos2 θ = 2
1
cos cos2 θ = 1
2. (d)
2
2
1 tan
cosec
+
=
2 2 2
2 2
2 2
sin cos sin1
cos cos1 1
sin sin
+ +
=
= 22
2
2
1sincos
1 cos
sin
=
= tan2 θ
3. (d)
2 2
1 1
1 tan 1 cos+
+ + =
2 2
1 1
sec cosec+
= cos2 θ + sin2 θ = 1
4. (a)
2 2
2 2
tan sec
cot cosec
−
− =
( )( )
2 2
2 2
sec tan
cosec cot
− −
− − =
1
1
−
−= 1
5. (c)
tan2 θ - sin2 θ = 2
2
sin
cos
- sin2 θ
= sin2 θ 2
11
cos
−
= sin2 θ
( )2
2
1 cos
cos
−
= sin2 θ 2
2
sin
cos
= sin2 θ tan2 θ
6. (b)
sin2A cos2B - cos2Asin2B
= sin2A (1-sin2B) - (1-sin2A) sin2B
= sin2A - sin2A sin2B - sin2B + sin2A sin2B
= sin2A - sin2B
7. (b)
(cosec2 θ - cot2 θ) = 1
⇒ 1+cot2 θ = cosec2 θ) =
2
2
cosec tan
sec
=
2
2
1 sin
cossin1
cos
=
2
2
sin cos cos
sinsin cos
=
= cot θ
8. (a)
(cosec2 θ - 1)(sec2 θ - 1)
= cot2 θ tan2 θ =
2 2
2 2
cos sin
sin cos = 1
9. (b)
cos2 θ + cos2 θ cot2 θ = cos2 θ (1 + cot2 θ)
= cos2 θ cosec2 θ = cos2 θ2
1
sin =
2
2
cos
sin
= cot2 θ
10. (d)
(cosec θ - sin θ)(sec θ - cos θ)(tan θ + cot θ)
= 1 1 sin cos
sin cossin cos cos sin
− − +
= 2 2 2 21 sin 1 cos sin cos
sin cos sin cos
− − +
=
2 2
2 2
cos sin 1
sin cos= 1
11. (d)
sec 1
sec 1
−
+ =
11
cos1
1cos
−
+
= −
+
1 cos
1 cos
= ( )( )
( )( )
− −
+ −
1 cos 1 cos
1 cos 1 cos =
( )2 2
2
1 cos 1 cos
sinsin
− − =
12. (c)
Solutions
Exercise - 1
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tan θ + cot θ = sin cos
cos sin
+
= 2 2sin cos 1
cos sin cos sin
+ =
= 1 1
cos sin
= secθ cosec θ
13. (c)
+ −
+− +
1 cos 1 cos
1 cos 1 cos
= ( )( )( )( )
1 cos 1 cos
1 cos 1 cos
+ +
− + + ( )( )
( )( )
1 cos 1 cos
1 cos 1 cos
− −
+ −
= ( ) ( )+ −
+− −
2 2
2 2
1 cos 1 cos
1 cos 1 cos
= 1 cos 1 cos
sin sin
+ − +
=
2
sin= 2 cosec θ
14. (d)
tan2A + cot2A = 2 2
2 2
sin A cos A
cos A sin A+
= 4 2
2 2
sin A cos A
sin Acos A
+
= ( )
22 2 2 2
2 2
sin A cos A 2sin Acos A
sin Acos A
+ −
= 2 2
2 2
1 2sin Acos A
sin Acos A
− = cosec2A sec2A – 2
15. (d)
1 1
sec 1 sec 1−
− + =
( )( )( )
− − −
+ −
sec 1 sec 1
sec 1 sec 1
= + − +
−2
sec 1 sec 1
sec 1 =
2
2
tan
[∵ sec2 θ - 1 = tan2 θ]
16. (a)
cot2A cosec2B - cot2B cosec2A
= cot2A (1+ cot2B) - cot2B (1 + cot2A)
= cot2A + cot2Acot2B - cot2B - cot2Acot2B
= cot2A - cot2B
17. (a)
tan sin
tan sin
+
− =
sinsin
cossin
sincos
+
−
=
11
sec 1cos1 sec 1
1cos
+ + = −
−
18. (b)
1 sin
1 sin
−
+
( )( )
1 sin
1 sin
−
−
= ( )
22
2 2
1 sin 1 sin 2sin
1 sin cos
− + − =
−
= sec2 θ + tan2 θ - 2 sec θ tan θ = (sec θ - tan θ)2
19. (d)
2 2
1 1
cos sin+
=
2 2
2 2 2 2
sin cos 1
cos sin cos sin
+ =
= sec2 θ cosec2 θ
20. (c)
sec4 θ - sec2 θ = sec2 θ (sec2 θ - 1) = (1 + tan2 θ) tan2 θ
= tan2 θ + tan4 θ = tan4 θ + tan2 θ
21. (d)
cos4 A + sin2 A cos2 A
= cos2 A(cos2 A + sin2 A) = cos2 A × 1 = cos2 A
22. (a)
sinA cosA sinA cosA
sinA cosA sinA cosA
+ −+
− +
( ) ( )( )( )
2 2sinA cosA sinA cosA
sinA cosA sinA cosA
+ + −
− +
( )
= =− − −
2 2 2 2
2 2
sin A cos A sin A 1 sin A=
2
2
2sin A 1−
23. (a)
2 2sec cosec+
= +
+ =
2 2
2 2 2 2
1 1 sin cos
cos sin sin cos
= 1
sin cos =
2 2sin cos
sin cos
+
= tan θ + cot θ
24. (d)
sin6A + cos6A = (sin2 A)3 + ( cos2 A)3
= (sin2A + cos2A)3 – 3sin2A cos2A (sin2A + cos2A)
= 1-3sin2Acos2A
25. (b)
cot A tanB
cot B tan A
+
+ =
1tanB
tan A1
tan AtanB
+
+
= ( )
( )
1 tanBtan A tanB
tan A 1 tan A tanB
+
+ =
tanB
tan A= cot A tan B
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Answer key 1 a 2 c 3 a 4 b 5 c
6 b 7 a 8 d 9 b 10 a
11 a 12 a 13 b 14 a 15 c
16 c 17 c 18 a 19 c 20 d
21 a 22 b 23 d 24 c 25 a
1. (a)
tan + cot = 2
⇒ tan +1
tan = 2 ⇒ tan2 + 1 = 2 tan
⇒ (tan - 1)2 = 0 ⇒ tan = 1
∴ cot = 1
tan5+cot5 = 15+15 = 2
2. (c)
sinn + cosecn ⇒ sin + 1
sin = 2
⇒ sin2+1 = 2sin ⇒ (sin−)2 = 0
⇒ sin = 1 ⇒ cosec = 1
[ Now sinn + cosecn = 1n+1n = 2 ]
3. (a)
cos + sec = 2
cos + 1
cos = 2 cos2 + = cos
⇒ (cos − ) = 1 cos =
∴ cos9 + sec9 = 1 + 1 = 2
4. (b)
tan + cot = 2 ⇒ tan + 1
tan = 2
tan2 + = tan ⇒ (tan - 1)2 = 0 tan =
Now tan-10 + cot-15 = 10 15
1 1
tan cos + =
1 1
1 1+ = 2
5. (c)
sin2 + sin2 = 2
Possible if and only if sin = sin = 1
∴ = = 90ᵒ
cos2
+
= cos90 90
2
+
= cos90ᵒ = 0
6. (b)
sec+tan = 2 ...... (i)
∴ sec - tan = 1
2 ......... (ii)
(i)+(ii) 2sec = 5
2 ⇒ sec =
5
4
7. (a)
cosec A + cot A = 3 ........ (i)
⇒ cosec A – cot A = 1
3 ........... (ii)
(i) + (ii)
2cosec A = 10
3 ⇒ cosec A =
5
3
5
3
4
A
cos A = 4
3
8. (d)
sec + tan = 5 + 2 ........... (i)
⇒ sec - tan = 1
5 2+ = 5 - 2 ..... (ii)
(i) + (ii)
2sec = 2 5 ⇒ sec = 5
Now
2
1
sin = 2
5, cos =
1
5
sin + cos = 2
5+
1
5 =
3
5
9. (b)
cosec = x+1
4x
Exercise - 2
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let cosec + cot = k ................. (i)
cosec - cot = 1
k .................. (ii)
(i) +(ii)
2 cosec = k+1
k ⇒ cosec =
1 1
2k
k
+
by comparison
1 1
2k
k
+
= x+
1
4x =
1 12
2 2x
x
+
⇒ k = 2x, 1
2x
cosec + cot = x
10. (a)
3sin +5cos = 5 ........... (i)
And let 5sin - 3cos = k .......... (ii)
(i)2 + (ii)2
32+52 = 52 + k2 ⇒ k = ±3
11. (a)
Let sin + cos = k ….(i)
sin - cos = 7
13 .... (ii)
(i)2+(ii)2
1+1 = k2+49
169 ⇒ k2 = 2-
49
169 =
289
169 ⇒ k = ±
17
13
12. (a)
sec2 + tan2 = 7 ⇒ (1 + tan2 ) + tan2 = 7
⇒ 2tan2 = 6 ⇒ tan2 = 3 ⇒ tan = 3
∴ = 60ᵒ
13. (b)
7 sin2+3cos2 = 4 ⇒ 7 sin2 + 3 (1-sin2) = 4
⇒ 7 sin2 + 3 – 3 sin2 = 4 ⇒ 4 sin2 = 1
⇒ sin2 =1
4 ⇒ sin =
1
2
= 30ᵒ
∴ tan 30ᵒ = 1
3
14. (a)
(cos+sin)2 + (cos-sin)2
= cos2 + sin2 + 2sincos + cos2+sin2 - 2sincos
= 2(sin2 + cos2) = 2
15. (c)
sec4 - tan4 = (sec2 - tan2) (sec2+tan2)
= 1×7
12 =
7
12
16. (c)
sec4 - sec2 = sec2(sec2−) = sec2 (tan2)
= (1+tan2)(tan2) = tan2 + tan4
17. (c)
x = a (sin + cos)
⇒ x
a = sin +cos ….. (i)
⇒ y = b (sin-cos)
⇒ y
b = sin - cos ..... (ii)
(i)2+(ii)2
2 2
2 2
x y
a b+ = (sin + cos)2 + (sin - cos)2
= sin2 + cos2 + 2sin cos + sin2 + cos2 -
2sin cos =
18. (a)
x2 + y2 + z2 = (sinAcosB)2+(sinAsinB)2+(cosA)2
= 2sin2Acos2B + 2sin2Asin2B + 2cos2A
= 2sin2A (cos2B+sin2B)+2cos2A
= sin2A + 2cos2A = 2 (sin2A+cos2A) = 2
19. (c)
cos = 1 – cos2
cos = sin2 = sin8 + 2sin6 + sin4
= cos4 + 2cos3 + cos2
= (cos2 + cos)2 = (1)2 = 1
20. (d)
cot + tan = x
cos sin
sin cos
+ = x ⇒
2 2cos sin
sin cos
+ = x
⇒ x = 1
sin cos and sec - cos = y
1
cos - cos = y ⇒
21 cos
cos
− = y ⇒ y =
2sin
cos
Now
( ) ( )2 3
2 23 2x y xy− =
2 32 43 2
2 2 2
1 sin 1 sin
cos sin cossin cos cos
−
= 1
= ( ) ( )2 2
3 33 3sec tan − = sec2 - tan2 = 1
21. (a)
tan - tan2 = 1
⇒ tan = 1+tan2 = sec2
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∴ sec2 - sec4 = tan - tan2 = 1
22. (b)
1 1
cos cot sinec −
−
= cosec + cot - cosec = cot
23. (d)
cosec - cot = 7
2 ............. (i)
then cosec+cot = 2
7 ........... (ii)
(ii)-(i)
2cot = 2
7-
7
2 =
4 49
14
− = -
45
14 ⇒ cot = -
45
28
24. (c)
cos2 - sin2 = 1
3
(cos4 - sin4)+1
= (cos2-sin2)(cos2+sin2)+1 = 1 4
13 3
+ =
25. (a)
cot180 (cot720 cos2220 +0 2 0
1
tan72 sec 68)
= cot180 (tan 180 cos2220 +0 2 0
1
cot18 cosec 22)
= cot180 (tan180 cos2220 + tan 180 sin2220)
= cot180 tan180 (cos2220 + sin2220) = 1 × 1 = 1
Answer key
1 a 2 b 3 b 4 c 5 c
6 b 7 b 8 a 9 c 10 a
11 a 12 b 13 d 14 a 15 a
16 d 17 b 18 c 19 a 20 b
21 d 22 c 23 d
1. (a)
tan sec 1
tan sec 1
+ −
− + =
( ) + −
− −
sec tan 1
1 sec tan
(Let sec θ + tan θ = x
then sec θ – tan θ = 1
x) =
− −=
−−
x 1 x 1
1 x 11
x x
= x
= sec θ + tan θ = 1 sin
cos cos
+
=
1 sin
cos
+
2. (b)
cot A cosecA 1
cot A cosecA 1
+ −
− + =
( )( )
cosecA cot A 1
1 cosecA cot A
+ −
− −
(Let cosec A + cot A = x
then cosec A - cot A = 1
x) =
x 1 x 1
1 x 11
x x
− −=
−−
= x
= cosec A + cot A = +1 cosA
sinA sinA =
1 cosA
sinA
+
3. (b)
sec A (1 – sin A) (sec A + tan A)
= sec A (1 - sin A) 1 sinA
cosA cosA
+
= sec A (1-sinA) ( )1 sin A
cos A
+=
2 2
2 2
1 sin A cos A
cos A cos A
−= = 1
4. (c)
2cosA sin A
1 tan A sin A cosA+
− − = +
−−
2cosA sin A
sinA sinA cosA1
cosA
= 2 2cos A sin A
cosA sin A sin A cosA+
− −
= 2 2sin A cos A
sin A cosA sin A cosA−
− −
Exercise - 3
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= 2 2sin A cos A
sin A cos A
−
−= sinA + cosA
5. (c)
2 3cos sin
1 tan sin cos
+
− − =
+
− −
2 3cos sin
sin sin cos1
cos
= 3 3cos sin
cos sin sin cos
+
− − =
−
− −
3 3sin cos
sin cos sin cos
= 3 3sin cos
sin cos
−
− =
( )( )( )
2 2sin cos sin cos sin cos
sin cos
− + +
−
= (1 + sin θ cos θ)
6. (b)
( )( )( )
+ + −
+
2 2sin cos sin cos sin cos
sin cos+sinθcos θ
= 1 - sin θ cos θ + sin θ cos θ = 1
7. (b)
sin2A + sec2A + 2sinAsecA + cos2A + cosec2A +
2cosAcosecA
= 1 + sec2A + cosec2A + 2sinA cosA
cosA sinA
+
= 1 + sec2A + cosec2A + 2 2 2sin A cos A
sin AcosA
+
= 1 + sec2A + cosec2A + 2sec A cosec A
= 1 + 2 2
1 1
cos A sin A+ + 2sec A cosec A
= 1 + 2 2
2 2
sin A cos A
sin Acos A
++ 2sec A cosec A
= 1 + sec2 A cosec2 A + 2sec A cosec A
= (1 + sec A cosec A)2
8. (a)
(1 + tanA tanB)2 + (tanA - tan B)2
= 1 + tan2A tan2B + 2tanA tanB + tan2A + tan2B - 2tan A
tan B
= 1 + tan2 A tan2 B + tan2A + tan2 B
= (1+tan2 A) (1+tan2B) = sec2 A sec2 B
9. (c)
sin θ (1 + tan θ) + cos θ (1 + cot θ)
= sin θsin
1cos
+
+ cos θ
cos1
sin
+
= sin θ cos sin
cos
+
+ cos θsin cos
sin
+
= (sin θ + cos θ) sin cos
cos sin
+
=(sin θ + cos θ)2 2sin cos
sin cos
+
= (sin θ + cos θ)1
sin cos
= sin cos
sin cos sin cos
+
=
1 1
cos sin+
= cosec θ + sec θ
10. (a)
= tan6 θ + 3 tan2 θ sec2θ + 1
= tan6 θ + 3 tan2 θ (1 + tan2 θ) + 1
= (tan2θ)3 + 3 tan2 θ (1 + tan2 θ) + 13
= (tan2 θ + 1)3 = (sec2 θ)3 = sec6 θ
11. (a)
2(sin6 θ + cos6 θ) - 3(sin4 θ + cos4 θ) + 1
= 2 [(sin2 θ + cos2 θ)3 - 3sin2 θ cos2 θ (sin2 θ + cos2 θ)] - 3
[(sin2 θ + cos2 θ)2 - 2sin2 θ cos2 θ] + 1
= 2 [1 - 3 sin2 θ cosec2 θ] - 3 (1 - 2sin2 θ cos2 θ) + 1
= 2 - 6 sin2 θ cos2 θ - 3 + 6 sin2 θ cos2 θ + 1 = 0
12. (b)
tan cot
sin cos
−
=
sin cos
cos sinsin cos
−
=
2 2
2 2
sin cos
sin cos
−
= 2 2
1 1
cos sin−
= sec2 θ - cosec2 θ
= 1 + tan2 θ - (1 + cot2 θ) = tan2 θ - cot2 θ
13. (d)
sec4A (1-sin4A) - 2 tan2A
= sec4A cos2A (1+sin2A) - 2tan2A
= sec4A 2
1
sec A(1+sin2A) - 2tan2A
= sec2A (1+sin2A) - 2tan2A
= sec2A + tan2A - 2 tan2A = sec2A - tan2A = 1
14. (a)
secA tanA
secA tanA
−
+ =
( )( )( )( )
secA tan A secA tan A
secA tan A secA tanA
− −
+ −
= ( )
2
2 2
sec A tan A
sec A tan A
−
−=
2 2sec A tan A 2secAtanA
1
+ −
= 1 + tan2 A + tan2 A - 2 sec A tanA
= 1 + 2 tan2 A - 2sec A tan A
= 1 - 2 sec A tan + 2 tan2 A
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5. (a)
2 sec2 θ – sec4 θ - 2 cosec2 θ + cosec4 θ
= 2 (1+tan2θ) - (1+tan2θ)2 - 2(1+cot2θ) + (1+cot2 θ)2
= 2 + 2 tan2 θ – (1 + tan4 θ + 2tan2 θ) - 2 - 2 cot2 θ + (1 +
cot4 θ + 2 cot2 θ) = cot4 θ - tan4 θ
16. (d)
− +
++ −
1 sin 1 sin
1 sin 1 sin
= ( )( )( )( )
− −
+ −
1 sin 1 sin
1 sin 1 sin+ ( )( )
( )( )
+ +
− +
1 sin 1 sin
1 sin 1 sin
= ( ) ( )− +
+− −
2 2
2 2
1 sin 1 sin
1 sin 1 sin
= ( ) ( )− +
+
2 2
2 2
1 sin 1 sin
cos cos
= 1 sin 1 sin
cos cos
− + +
= (sec θ - tan θ) + (sec θ + tan θ)
= 2 sec θ
17. (b)
2 cosec2230 cot2670 - sin2230 - sin2670 - cot2670
= (2 cosec2230 - 1) cot2670 - sin2230 - sin2670
= cot2670 [cosec2230 + cosec2230 - 1] - (sin2230 +
cos2230)
= tan 2230 [cosec2230 + cot2230] - 1
= tan2230 × cosec2230 + tan2230cot2230 -1
= 2 0
2 0 2 0
sin 23 1
cos 23 sin 23+ 1 – 1 =
2 0
1
cos 23= sec2230
18. (c)
sin cos
cos sin1 1
sin cos
+
− −
= 2 2sin cos
sin cos cos sin
+
− −
= 2 2sin cos
sin cos sin cos
−
− −
= 2 2sin cos
sin cos
−
− = sin θ + cos θ
19. (a)
cos cos
1 sin 1 sin
+
− +
=( )
( )( )( )
( )( )
cos 1 sin cos 1 sin
1 sin 1 sin 1 sin 1 sin
+ − +
− + + −
= ( ) ( )
2 2
cos 1 sin cos 1 sin
1 sin 1 sin
+ − +
− −
= 1 sin 1 sin
cos cos
+ − +
=
1 sin 1 sin
cos
+ + −
=2
cos= 2 sec θ
20. (b)
sinA 1 cosA
1 cosA sinA
++
+ =
( )( )
22sin A 1 cosA
sinA 1 cosA
+ +
+
= ( )
2 2sin A 1 cos A 2cosA
sinA 1 cosA
+ + +
+
= ( )
2 2sin A cos A 1 2cosA
sinA 1 cosA
+ + +
+
= ( ) ( )
1 1 2cosA 2 2cosA
sinA 1 cosA sinA 1 cosA
+ + +=
+ +
= ( )
( )
2 1 cosA
sin A 1 cosA
+
+ =
2
sinA= 2cosec A
21. (d)
(1 + cot θ - cosec θ)(1 + tan θ + sec θ)
= cos 1 sin 1
1 1sin sin cos cos
+ − + +
= sin cos 1 cos sin 1
sin cos
+ − + +
= ( )
2 2sin cos 1
sin cos
+ −
=
2 2sin cos 2sin cos 1
sin cos
+ + −
= 1 2sin cos 1
sin cos
+ −
=
2sin cos
sin cos
= 2
22. (c)
tan tan
sec 1 sec 1
−
− + = tan θ
1 1
sec 1 sec 1
+
− +
= tan θ( )( )sec 1 sec 1
cos 1 sec 1
+ + −
− +
=2sec 2sec 2sec cos
sintan sin
cos
= =
= 2
sin [∵ sec θ cos θ = 1]
= 2 cosec θ
23. (d)
tan2 𝜙 + cot2 𝜙 + 2
= tan2 𝜙 + cot2 𝜙 + 2 tan 𝜙 × cot 𝜙 [∵ tan 𝜙 cot 𝜙 = 1]
= (tan 𝜙 + cot 𝜙)2 =
2sin cos
cos sin
+
=
2 22sin cos 1
cos sin cos sin
+ =
=
21 1
cos sin
= (sec 𝜙 cosec𝜙)2
= sec2 𝜙 cosec2 𝜙
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Trigonometry
Sum and difference of angles
of T-Ratio
Chapter - 06
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* sin (A + B) = sin A cos B + cos A sin B
* sin (A – B) = sin A cos B – cos A sin B
* cos (A + B) = cos A cos B – sin A sin B
* cos (A – B) = cos A cos B + sin A sin B
* tan (A + B) = ( )( )
sin A B
cos A B
+
+
= sinAcosB cosAsinB
cosAcosB sinAsinB
+
−
= tanA tanB
1 tanAtanB
+
−
* tan (A – B) = tanA tanB
1 tanAtanB
−
+
* cot (A + B) = cot Acot B 1
cot B cot A
−
+
* cot (A – B) = cot Acot B 1
cot B cot A
+
−
* sin (A + B) sin (A – B)
= (sin A cos B + cos A sin B) (sin A cos B – cos A sin B)
= sin2 A cos2 B – cos2 A sin2 B
= sin2 A (1 – sin2 B) – (1 – sin2 A) sin2 B
= sin2 A – sin2 A sin2 B – sin2 B + sin2 A sin2 B
= sin2 A – sin2 B
= 1 – cos A –( A– cos2 B)
= cos2 B–cos2 A
*sin (A + B) sin (A – B) = sin2 A – sin2 B = cos2 B – cos2 A
* cos (A + B) cos (A – B) = cos2 B – sin2 A = cos2 A - sin2 B
* tan (A + B + C)
= tan [(A + B) + C]
=
++
−
+ −
−
tan A tanBtanC
1 tan A tanBtan A tanB
1 tanC1 tan A tanB
= + + −
− − −
tanA tanB tanC tanAtanBtanC
1 tanAtanB tanBtanC tanCtanA
* Values at 15°, 75°
sin15°
= sin(45° – 30°)
= sin45° cos30° – cos45°sin30°
= 1 3 1
2 22 2
− = 3 1
2 2
−
cos15°
= cos(45° – 30°)
= cos45° cos30° – sin45°sin30°
= + 1 3 1 1
2 22 2 = 3 1
2 2
+
sin75°
= sin(90° - 15°) = cos15° = 3 1
2 2
+
cos75°
= cos(90° - 15°) = sin15° = −3 1
2 2
sin15° = cos75° = −3 1
2 2
sin75° = cos15° = 3 1
2 2
+
tan15°
= tan (45° – 30°) = tan45 tan30
1 tan45 tan30
−
+
= ( )( )
( )( )
− − −=
+ −+
11 3 1 3 13
1 3 1 3 113
= 3 1 2 3
2 32
+ −= −
tan75°
= tan(45° + 30°) = tan45 tan30
1 tan45 tan30
+
−
=
13
13
+
−
= ( )( )
3 13 1
3 1 3 1
++
− − =
3 1 2 32 3
2
+ += +
cot15° = cot(90° – 75°) = tan75° = 2 3+
cot75° = cot(90° – 15°) = tan15° = 2 3−
tan15° = cot75° = 2 3−
tan75° = cot15° = 2 3+
Sum and Difference of angles of
T - Ratio 6
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1 Value of ( )sin A B
cosAcosB
− =?
Sol. ( )−sin A B
cosAcosB =
−sinAcosB cosAsinB
cosAcosB
= sinAcosB cosAsinB
cosAcosB cosAcosB− = tan A – tan B
2 If sin A = 3
5 and cos B =
12
13 where (0° A, B 90°) then
find value of
(i) sin (A + B) (ii) cos (A + B)
Sol. sin A = 3
5
cos A =
23 4
15 5
− =
cos B = 12
13
sin B =
212 5
113 13
− =
sin (A + B) = sin A cos B + cos A sin B
= 3 12 4 5
5 13 5 13 + =
36 20 56
65 65 65+ =
cos (A + B) = cos A cos B – sin A sin B
= 4 12 3 5
5 13 5 13 − =
33
65
3 If (A + B) = 45° then
(i) tan A + tan B + tan A tan B = ?
(ii) (1 + tan A) (1 + tan B) =?
Sol. (i) A + B = 45°
tan (A + B) = tan45° tanA tanB
1 tanAtanB
+
− = 1
tan A + tan B = 1 – tan A tan B
tan A + tan B + tan A tan B = 1
(ii) tan A + tan B + tan A tan B = 1
1 + tan A + tan B + tan A tan B = 1 + 1
(1 + tan A) + tan B(1 + tan A) = 2
(1 + tan A)(1 + tan B) = 2
4 If A + B = 45° then (1 – cot A)(1 – cot B) = ?
Sol. A + B = 45°
cot (A + B) = cot 45° cot Acot B 1
cot A cot B
−
+ = 1
cot A cot B – 1 = cot A + cot B
1 + cot A cot B – cot A – cot B = 1 + 1
1 – cot A – cot B + cot A cot B = 2
1 – cot A – cot B(1 – cot A) = 2
(1 – cot A)(1 – cot B) = 2
5 If tan x tan y = 1
3 then 2 cos (x + y) =?
Sol. tan x tan y = 1
3
sinx siny 1
cosx cosy 3=
Using componendo and dividendo
cosxcosy sinxsiny 3 1
cosxcosy sinxsiny 3 1
− −=
+ +
( )( )
cos x y 1
2cos x y
+=
− 2 cos (x + y) = cos (x – y)
6 If x = sec2 + cosec2 and y = cot2 + tan2 then 1 + x –
y = ?
Sol. 1 + x – y
= 1 + sec2 + cosec2 – cot2 + tan2 = 1 + 1 + 1 = 3
7 sin 15° + cos 15° = ?
Sol sin 15° + cos 15°
= 3 1 3 1
2 2 2 2
− ++ =
3 1 3 1
2 2
− + + =
3
2
Examples
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1. tan (45°+A) =?
(a) 1 tan
1 tan
A
A
+
− (b) 1
(c) 1 tan
1 tan
A
A
−
+ (d)
2tan
1 tan
A
A+
2. cot(45° - A) =?
(a) 1 cot
1 cot
A
A
+
− (b)
cot 1
cot 1
A
A
−
+
(c) cot 1
cot 1
A
A
+
− (d) 1
3. ( )( )
tan tan
1 tan tan
A B B
A B B
+ −
+ −=?
(a) tan (A+2B) (b) tan A
(c) tan B (d) 1
4. tan30 tan15
tan30 tan15 1
+
−
(a) 1 (b) -1 (c) 0 (d) tan 150
5. Find tan (A + B) if tan A = 1
2 and tan B =
1
3
(a) 5
6 (b) -1 (c)
6
5 (d) 1
6. sin22 cos22
sin22 cos22
+
− =?
(a) tan 67° (b) – tan 67°
(c) tan 23° (d) – tan 23°
7. tan69 tan66
11 tan69 tan66
+ +
−
(a) 2 (b) 1 (c) -1 (d) 0
8. sin (45° + A) – cos (45° – A) = ?
(a) 1
2 (b) 1 (c) 0 (d)
3
2
9. cos (600 + A) + cos (60° - A) =?
(a) 1
2 (b) sin A (c) cos A (d) 3
10. 7 7
sin cos cos sin12 4 12 4
− =?
(a) 1 (b) 1
2 (c)
3
2 (d) 0
11. cos2 x4
+
– sin2 x
4
−
= ?
(a) cos2x
2 (b) cos 2x (c)
3
2sin 2x (d) 0
12. cosA + cos (120° - A) + cos (120° + A) =?
(a) 0 (b) 2cos A (c) - 1 (d) 1
13. cot 4
A
+
cot 4
A
−
=?
(a) - 1 (b) 1 (c) 0 (d) cot A
14. 2 sin (x – 45°) = ?
(a) sin x + cos x (b) sin x - cosx
(c) 2 (sin x – cos x) (d) cos x - sin x
15. 2 cos 4
x
+
=?
(a) sin x + cos x (b) sin x - cosx
(c) 2 (sin x – cos x) (d) cos x - sin x
16. 1 tan tan
1 tan tan
A B
A B
−
+= ?
(a) ( )( )
cos
cos
A B
A B
+
− (b) ( )
( )
cos
cos
A B
A B
−
+
(c) ( )( )
tan
tan
A B
A B
+
− (d) ( )
( )
tan
tan
A B
A B
−
+
17. =−
+
17sin17cos
17sin17cos
(a) 62tan (b) 56tan (c) 54tan (d) 73tan
18. =−
+
9sin9cos
9sin9cos
(a) cot 360 (b) 36tan (c) 18tan (d) None of these
Exercise - 1
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19. =− 75cot75tan
(a) 32 (b) 32 + (c) 32 − (d) None of these
20. If ,sinsincoscos)cos( BABABA +=+ then ),( =
(a) (– 1, – 1) (b) (– 1, 1) (c) (1, – 1) (d) (1, 1)
21. tan 16° + tan 29° + tan 16° tan 29° = ?
(a) 1 (b) -1 (c) 0 (d) 2
22. ( )( )
− +
+ +
tan2 tan
1 tan2 tan =?
(a) tan ( + ) (b) –tan (2 – )
(c) tan (2 – ) (d) tan ( – )
23. If 5
3)cos( =− BA and ,2tantan =BA then
(a) 1
cos cos5
A B = −
(b) 2
sin sin5
A B =
(c) 5
1coscos −=BA
(d) 5
1sinsin −=BA
24. =
+
+
−
147cos
147sin
12sin12cos
12sin12cos
(a) 1 (b) – 1 (c) 0 (d) None of these
25. tan3 tan2 tanA A A− − =?
(a) AAA tan2tan3tan
(b) AAA tan2tan3tan−
(c) AAAAAA tan3tan3tan2tan2tantan −−
(d) None of these
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1. 3
tan tan4 4
+ +
=?
(a) 1 (b) -1 (c) 2tan (d)
2. If A + B = 45° and tan A = m + 1 then find the value of
tan B in terms of m.
(a) m – 1 (b) m
m 1+ (c)
m
m 2
−
+ (d) 1 – m2
3. Find sin(A + B), if sinA = 3
5 and cos B =
5
13
(a) 48
65 (b)
63
65 (c)
12
13 (d) 1
4. Find sin (A -B) if cosA = 12
13 and cosB =
8
17
(a) 40
121− (b)
180
221− (c) −
140
221 (d)
180
221
5. cos (A - B) cos(A - B) =?
(a) cos2 A - sin2 B (b) cos2 B - sin2 A
(c) sin2 A - sin2 B (d) cos2 A - cos2 B
6. ( ) ( )( ) ( )
sin sin sin
cos cos cos
A B A A B
A B A A B
− + + +
− + + + =?
(a) sin (2A + B) (b) tan A
(c) cot B (d) tan B
7.
− + + tan cot
4 4A A =?
(a) 1 (b) 2 tan A (c) 2 cot A (d) 0
8. sin(A - B) sin(A + B) + sin (B - C) sin (B + C) + sin (C - A)
sin (C + A) = 0
(a) sin2 A – sin2 B – sin2C (b) sin2 A sin2 B sin2C
(c) 1 (d) 0
9. If 10
1sin =A and ,
5
1sin =B where A and B are
positive acute angles, then =+ BA
(a) (b) 2/ (c) 3/ (d) 4/
10. (1+tan 1) (1 + tan 2) (1 + tan3) …… (1 + tan 45°) = ?
(a) 221 (b) 222 (c) 223 (d) 224
11. (1-cot6°) (1-cot7°) (1 – cot7°) (1 – cot8°) …. (1 –cot39°)
= 2x, x = ?
(a) 15 (b) 16 (c) 17 (d) 18
12. If 7
1cos =P and ,
14
13cos =Q where P and Q both are acute
angles. Then the value of QP − is
(a) o30 (b) o60 (c) o45 (d) o75
13. sin163 cos347 sin73 sin167o o o o+ =?
(a) 0 (b) 1/2 (c) 1 (d) None of these
14. The value of + 150sin120cos330cos600sin is
(a) – 1 (b) 1 (c) 2
1 (d)
2
3
15. tan15 tan20 tan25 tan15 tan20 tan25
1 tan15 tan20 tan20 tan25 tan25 tan15
+ + −
− − − = ?
(a) 1 (b) 3 (c) 3− (d) 1
3
16. If tan ( + ) = 3
4 and tan ( – ) =
5
12 then find the
value of tan 2.
(a) 16
33 (b)
16
63 (c)
56
63 (d)
16
33
17. ( )sin
cos cos
A B
A B
−+
( )sin
cos cos
B C
B C
− +
( )sin
cos cos
C A
C A
− =?
(a) 2(tan A + tan B + tan C) (b) 0
(c) 1 (d) None of these
18. 2 ( cos 105° + sin 105°) = ?
(a) 0 (b) -1 (c) 1 (d) 3 1
2 2
−
19. If ,225=+ BA then( ) ( )
cot cot
1 cot 1 cot=
+ +
A B
A B?
(a) 1 (b) – 1 (c) 0 (d) ½
Exercise - 2
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20. If tan,tan are the roots of the equation
),0( 02 =++ pqpxx then
(a) p−=+ )sin( (b) 1
)tan(−
=+q
p
(c) q−=+ 1)cos( (d) None of these
21.
+ + − tan tan
4 4A A =?
(a) cos 2 A (b) sec 2A (c) 2 sec 2A (d) 2 cos 2A
22. If ,cottan2tan BBA += then =− )tan(2 BA
(a) Btan (b) Btan2 (c) Bcot (d) Bcot2
23. =
−−
+
6sin
6cos 22
(a) 2cos2
1 (b) 0 (c) 2cos
2
1− (d)
2
1
24. If cot (A + B) = 4
3 and cot (A – B) =
12
5 then find the
value of tan 2B
(a) 16
63 (b)
56
63 (c)
16
33 (d)
56
33
25. sin (n + 1)A sin (n + 2)A + cos (n + 1)A cos (n + 2)A = ?
(a) cos (2n + 3)A (b) cos A
(c) –cos A (d) –cos (2n + 3)
26. if 2
1tan −=A and ,
3
1tan −=B then =+ BA
(a) 4
(b)
4
3 (c)
4
5 (d) None of these
27. If 5
4sin =A and ,
13
12cos −=B where A and B lie in first and
third quadrant respectively, then =+ )cos( BA
(a) 65
56 (b)
65
56− (c)
65
16 (d) –
65
16
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Answer key
1 a 2 c 3 b 4 b 5 d
6 a 7 d 8 c 9 c 10 c
11 d 12 a 13 b 14 b 15 d
16 a 17 a 18 a 19 a 20 c
21 a 22 d 23 a 24 c 25 a
1. (a)
tan (45° - A) = tan45 tan
1 tan45 tan
A
A
+
− =
1 tan
1 1 tan
A
A
+
−
= 1 tan
1 tan
A
A
+
−
2. (c)
cot (45°-A) = cot cot 45 1
cot cot 45
A
A
+
−
= cot 1 1
cot 1
A
A
+
− =
cot 1
cot 1
A
A
+
−
3. (b)
( )
( )
tan tan
1 tan tan
A B B
A B B
+ −
+ +
= tan [(A + B) – B] = tan [A + B - B] = tan A
4. (b)
tan30 tan15
tan30 tan15 1
+
− =
+ −
−
tan30 tan15
1 tan30 tan15
= - tan (30°+ 15°) = - tan 45° = -1
5. (d)
tan A = 1
2 and tan B =
1
3
now tan (A + B) = tan tan
1 tan tan
A B
A B
+
−
=
1 1
2 31 1
12 3
+
−
=
5
65
6
= 1
6. (a)
sin22 cos22
sin22 cos22
+
− =
tan22 1
tan22 1
+
−
= 1 tan22
1 tan22
+ −
− = -tan (45° + 22°) = -tan 67°
7. (d)
tan69 tan66
1 tan69 tan66
+
− + 1
= tan (69° + 66°) + 1 = tan135° + 1 = -1 + 1 = 0
8. (c)
(sin 45° cos A + cos 45° sin A) – (cos 45° cos A + sin 45°
sin A)
= 1 1 1 1
cosA sin A cosA sin A2 2 2 2
+ − − = 0
9. (c)
cos(60°+A) + cos(60°-A)
= (cos60° cosA – sin 60° sin A) + (cos60° cosA + sin60°
sin A)
= 2 cos 60° cos A = 1
22
cosA = cos A
10. (c)
7 7
sin cos cos .sin12 4 12 4
−
= 7sin
12 4
−
=
3sin
3 2
=
11. (d)
Using cos (A + B)cos (A – B) = cos2 A – sin2 B
= + + − + − +
cos x x cos x x4 4 4 4
=
cos cos2x2
= 0
12. (a)
cosA + (cos120° cosA + sin 1200 sin A) + (cos120° cosA
– sin 120° sin A)
= cos A + 2 cos120° cosA = cosA +
−
12
2 cosA
= cosA – cos A = 0
Solutions
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13. (b)
cot cot4 4
A A
+ −
= cot cot 1 cot cot 1
4 4
cot cot cot cot4 4
A A
A A
− +
+ −
= − +
+ −
cot 1 cot 1
cot 1 cot 1
A A
A A = 1
14. (b)
2sin (x -45°)
= 2 (sinx cos45° - sin450cosx)
= 2
−
1 1sin cos
2 2x x
= 2
2(sinx - cosx) = sinx – cosx
15. (d)
cos 4
x
+
= 2 × cos cos sin sin4 4
x x
−
= 2 1 1
cos sin2 2
x x
−
= cosx – sinx
16. (a)
1 tan tan
1 tan tan
A B
A B
−
+ =
sin sin1 .
cos cossin sin
1 .cos cos
A B
A BA B
A B
−
+
=
cos cos sin sin
cos coscos cos sin sin
cos cos
A B A B
A BA B A B
A B
−
+=
( )( )
cos
cos
A B
A B
+
−
17. (a)
oo
oo
17sin17cos
17sin17cos
−
+
Divided by o17cos in numerator and denominator,
1 tan 17
1 tan 17
o
o
+=
− = tan 45 tan 17
1 tan 45 tan 17
o o
o o
+
−
= ( )tan 45 17+ = tan620
18. (a)
cos 9 sin 9
cos9 sin 9
o o
o o
+
− =
o
o
9tan1
9tan1
−
+
= ( )tan 45 9o o+ = tan54o = tan (900 - 360) = cot 360
19. (a)
tan75 cot75o o− = tan75 tan15o o−
32)32()32( =−−+=
20. (c)
BABABA sinsincoscos)(cos +=+
But BABABA sinsincoscos)(cos −=+ 1, 1 = = −
21. (a)
tan (16° + 29°) = tan16 tan29
1 tan16 tan29
+
−
tan 45° = 1 = tan16 tan29
1 tan16 tan29
+
−
1 – tan16° tan29° = tan16° + tan29°
tan16° + tan29° + tan16° tan29° = 1
22. (d)
( )( )
tan2 tan
1 tan2 .tan
− +
+ + = tan (2 – – ) = tan ( – )
23. (a)
5
3)(cos =− BA
3sinsin5coscos5 =+ BABA …..(i)
From 2nd relation,
BABA coscos2sinsin = .....(ii)
5
1coscos =BA and 5sin sin 2A B =
24. (c)
o
o
oo
oo
147cos
147sin
12sin12cos
12sin12cos+
+
−
o
o
o
147tan12tan1
12tan1+
+
−=
( ) ( )tan 45 12 tan 180 33o o o o= − + −
= tan 330 – tan 330 = 0
25. (a)
( )tan tan 2
tan 3 tan 21 tan tan 2
A AA A A
A A
+= + =
−
tan 3 tan 3 tan 2 tan tan tan 2A A A A A A − = +
AAAAAA tan2tan3tantan2tan3tan =−−
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Answer key
1 b 2 c 3 b 4 c 5 a
6 b 7 d 8 d 9 d 10 c
11 c 12 b 13 b 14 a 15 b
16 d 17 b 18 c 19 d 20 b
21 c 22 c 23 a 24 a 25 b
26 b 27 d
1. (b)
tan3
tan4 4
+ +
=
+
−
tan tan4
1 tan tan4
+
−
3tan tan
43
1 tan tan4
= 1 tan
1 1 tan
+
− ×
( )
− +
− −
1 tan
1 1 tan
= ( )( )
+
−
1 tan
1 tan×
( )( )
tan 1
1 tan
−
+ =
( )( )
tan 1
1 tan
−
−
=( )( )
−−
−
1 tan
1 tan = -1
2. (c)
tan (A + B) = tanA tanB
1 tanAtanB
+
−
1 = ( )
m 1 tanB
1 m 1 tanB
+ +
− +
m + 1 + tan B = 1 – m tan B – tan B
(2 + m) tan B = -m tan B = m
m 2
−
+ 3. (b)
sin A = 3
5
cosA = 21 sin A− =
23
15
−
=
16
25 =
4
5
and cos B = 5
13so
sinB = 21 cos B− =
25
113
−
=
144
169 =
12
13
Now sin(A + B) = sin A cosB + cos A sinB
=3 5 4 12
5 13 3 13 + =
15 48 15 48
65 65 65
++ = =
63
65
4. (c)
cos A = 12
13
sin A = 21 cos A− =
212
113
−
=
5
13
and cos B = 8
17
sin B = 21 cos B− =
28
117
−
=
15
17
Now, sin (A - B) = sin A cosB – cosA sin B
= 5 8 12 15
13 17 13 17 − = −
40 180
221 221= −
140
221
5. (a)
cos(A + B) cos(A-B)
= (cos A cos B – sin A sin A sin B) (cos A cos B + sin A
sinB)
= cos2 A cos2 B –sin2A sin2B
= cos2 Acos2B- (1- cos2A) (1- cos2B)
= cos2A cos2B- (1- cos2B-cos2A+cos2Acos2B)
= cos2Acos2B-1+cos2B + cos2 A – cos2Acos2B
= cos2A -1 + cos2 B = cos2 A – (1-cos2B)
= cos2 A - sin2 B
6. (b)
( ) ( )( ) ( )
sin sin sin
cos cos cos
A B A A B
A B A A B
− + + +
− + + +
( ) ( )( ) ( )
− + + +=
+ + + −
sin cos cos sin sin sin cos cos sin
cos cos sin sin cos cos cos sin sin
A B A B A A B A B
A B A B A A B A B
= sin 2sin cos
cos 2cos cos
A A B
A A B
+
+=
( )sin 1 2cos
cos (1 2cos )
A B
A B
+
+
= sin
cos
A
A = tan A
7. (d)
tan cot4 4
A A
− + +
= tan tan cot cot 1
4 4
1 tan tan cot cot4 4
A A
A A
− −
+
+ +
Exercise - 2
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71
= − −
++ +
tan 1 cot 1
1 tan 1 cot
A A
A A =
−−
++
+
11
tan 1 tan11 tan
1tan
A AA
A
= − −
++ +
tan 1 1 tan
1 tan tan 1
A A
A A=
− −−
+ +
tan 1 tan 1
1 tan tan 1
A A
A A = 0
8. (d)
sin (A - B) sin (A + B) + sin (B - C) sin (B + C) + sin (C - A)
sin (C + A)
= sin2A – sin2B + sin2B – sin2C + sin2C – sin2 A
[ sin (A - B) sin (A + B) = sin2 A – sin2 B] = 0
9. (d)
BABABA sincoscossin)(sin +=+
10
11
5
1
5
11
10
1−+−=
1 4 1 9
5 1010 5= + =
1 5 1(2 3)
50 50 2+ = =
4
sin)(sin
=+ BA Hence, 4
=+ BA
10. (c)
(1+ tan1°) (1 + tan 2°) (1 + tan 3°) (1 +tan 4°) ……………
(1 + tan 45°)
= ( 1+ tan 1°) (1+ tan 44°) (1 +tan2°) (1 + tan 43°)
………22 times (1 + tan45°)
= 2 × 2 ………………… 22 times × (1 × 1)
= 222 × 2 = 223
11. (c)
( 1 – cot6°) (1-cot39°) (1- cot7°) (1 – cot38°)…… 34
terms
= [(1 – cot6°) (1 – cot39°)] [(1- cot7°) (1 – cot38°)]
……….. 17 terms
= [1- cot6° - cot39° + cot6° cot31°] [(1cot7° - cot39° +
cot7° cot39°)] ……………… 17terms
= 2 × 2 x ……. 17 times = 217
x = 17
12. (b)
14
13cos,
7
1cos == QP
QPQPQP sinsincoscos)cos( +=−
1 13 48 27
7 14 7 14= +
= 13 36 1
cos6098 2
o+= = oQP 60=−
13. (b)
oooo 167sin73sin347cos163sin +
)1790cos()13360cos()17180sin( oooooo −+−−=
)13180sin( oo −
sin17 cos13 cos17 sin13o o o o= +
( )sin 17 13 sin30 1/ 2o o= + = =
14. (a)
oooo 150sin120cos330cos600sin +
oooo 60cos30sin30cos60sin −−=
sin (60 30 )o o= − + = sin90 1− = −
15. (b)
tan15 tan20 tan25 tan15 tan20 tan25
1 tan15 tan20 tan20 tan25 tan25 tan15
+ + −
− − −
= tan (150 + 200 + 250) = tan 60° = 3
16. (d)
tan 2 = tan [( + ) + ( – )]
= ( ) ( )
( ) ( )
tan tan
1 tan tan
+ + −
− + − =
3 5
4 123 5
14 12
+
−
= 56
33
17. (b)
( )−sin
cos cos
A B
A B=
−sin cos cos sin
cos cos
A B A B
A B=tanA –tanB
( )−sin
cos cos
B C
B C=
−sin cos cos sin
cos cos
B C B C
B C= tanB-tanC
( )−sin
cos cos
C A
C A=
sin cos cos sin
cos cos
C A C A
C A
−= tanC-tanA
Now
( )sin
cos cos
A B
A B
−+
( )sin
cos cos
B C
B C
− +
( )sin
cos cos
C A
C A
−
= tan A – tan B + tan B –tan C + tan C –tan A = 0
18. (c)
( )2 cos105 sin105 +
= ( ) ( )2 cos 60 45 sin 60 45 + + +
= 2 [cos60°cos45°-sin60°sin45°+sin60°cos45° +
cos60°sin45°]
= 21 1 3 1 3 1 1 1
2 2 2 22 2 2 2
− + +
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= 2 1 3 3 1
2 2 2 2 2 2 2 2
− + +
= 1 1
22 2 2 2
+
=
2 2
2 2 = 1
19. (d)
A + B = 2250 tan( ) tan 225+ = oA B
tan tan 1 tan tanA B A B + = −
(1+tanA)(1+tanB) = 2
1
1cot A
+
11
cot B
+
= 2
( ) ( )
cot cot
1 cot 1 cot+ +
A B
A B=
1
2
20. ( b)
Since tan,tan are the roots of the equation
2 0x px q+ + =
,tantan p−=+ q= tantan
tan tan
tan ( )1 tan tan
++ =
− ( )1 1
p p
q q
−= =
− − −
21. (c)
= tan4
A
+
+tan4
A
−
=+ −
+− +
1 tan 1 tan
1 tan 1 tan
A A
A A
( ) ( )
( )( )
+ + −
− +
2 21 tan 1 tan
1 tan 1 tan
A A
A A=
( )( )
+
−
2
2
2 1 tan
1 tan
A
A
=
+
−
2
2
2
2
sin2 1
cos
sin1
cos
A
A
A
A
= ( )+
−
2 2
2 2
2 cos sin
cos sin
A A
A A=
2 1
cos2A
= 2 sec
2A
22. (c)
,cottan2tan BBA +=
+
−=−
BA
BABA
tantan1
tantan2)(tan2
= (2 tan cot tan )
21 (2 tan cot ) tan
B B B
B B B
+ −
+ +
= 2
tan cot2
2(1 tan )
B B
B
+
+ B
B
BBcot
)tan1(
)1(tancot2
2
=+
+=
23. (a)
−−
+
6sin
6cos 22
+−+
−++=
66cos
66cos
)]cos()cos(sincos[ 22 BABABA −+=−
1
cos cos2 cos23 2
= =
24. (a)
tan 2B = ( ) ( )
1 1
cot2B cot A B A B=
+ − −
= ( ) ( )
( ) ( )
cot A B cot A B
cot A B cot A B 1
− − +
− + + =
12 4 16165 3 15
12 4 63 631
5 3 15
−
= =
+
25. (b)
cos x cos y + sin x sin y = cos (x – y)
sin (n + 1)A sin (n + 2)A + cos (n + 1)A cos (n + 2)A
= cos {(n + 1)A – (n + 2)A}
= cos(n+1–n–2)A = cos (-A) = cos A
26. (b)
2
1tan −=A and
3
1tan −=B
1 1tan tan 2 3tan( ) 1
1 11 tan tan1
2 3
A BA B
A B
− −+
+ = = = −−
−
3
tan ( ) tan4
A B
+ =
Hence, 3
4A B
+ =
27. (d)
5
4sin =A and
13
12cos −=B
BABABA sinsincoscos)(cos −=+
169
1441
5
4
13
12
25
161 −−
−−=
65
16
13
5
5
4
13
12
5
3−=
−−−=
(Since A lies in first quadrant and B lies in third
quadrant)
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Trigonometry
Transformation of a Product
into a sum or Difference
Chapter - 07
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TRANSFORMATION OF A
PRODUCT INTO A SUM OR DIFFERENCE
(i) 2 sin A cos B = sin (A + B) + sin (A – B)
(ii) 2 cos A sin B = sin (A + B) – sin (A – B)
(iii) 2 cos A cos B = cos (A + B) + cos (A – B)
(iv) 2 sin A sin B = cos (A – B) – cos (A + B)
(i) sin (A + B) + sin (A – B)
= (sin A cos B + cos A sin B) + (sin A cos B – cos A sin B)
= 2 sin A cos B
2 sin A cos B = sin (A + B) + sin (A – B)
(ii) sin (A + B) – sin (A – B)
= (sin A cos B + cos A sin B) – (sin A cos B – cos A sin B)
= 2 cos A sin B
2 cos A sin B = sin (A + B) – sin (A – B)
(iii) cos (A + B) + cos (A – B)
= (cos A cos B – sin A sin B) + (cos A cos B + sin A sin B)
= 2 cos A cos B
2 cos A cos B = cos (A + B) + cos (A – B)
(iv) cos (A – B) – cos (A + B)
= cos A cos B + sin A sin B) – (cos A cos B – sin A sin B)
= 2 sin A sin B
2 sin A sin B = cos (A – B) – cos (A + B)
TRANSFORMATION OF A
SUM OR DIFFERENCE INTO A PRODUCT
(i) sin C + sin D = 2 sin C D C D
cos2 2
+ −
(ii) sin C – sin D = 2 cos C D C D
sin2 2
+ −
(iii) cos C + cos D = 2 cos C D C D
cos2 2
+ −
(iv) cos C – cos D = 2 sin C D D C
sin2 2
+ −
(i) sin C + sin D
= sin (A + B) + sin (A – B)
= (sin A cos B + cos A sin B) + (sin A cos B – cos A sin B)
= 2 sin A cos B = 2 sin C D C D
cos2 2
+ −
sin C + sin D = 2 sin C D C D
cos2 2
+ −
(ii) sin C – sin D
= sin (A + B) – sin (A – B)
= (sin A cos B + cos A sin B) – (sin A cos B – cos A sin B)
= 2 cos A sin B = 2 cos C D C D
sin2 2
+ −
sin C – sin D = 2 cos C D C D
sin2 2
+ −
(iii) cos C + cos D
= cos (A + B) + cos (A – B)
= (cos A cos B – sin A sin B) + (cos A cos B + sin A sin B)
= 2 cos A cos B = 2 cos C D C D
cos2 2
+ −
cos C + cos D = 2 cos C D C D
cos2 2
+ −
(iv) cos C – cos D
= cos (A + B) – cos (A – B)
= (cos A cos B – sin A sin B) – (cos A cos B + sin A sin B)
= – 2 sin A sin B = 2 sin A (– sin B)
= 2 sin A sin (– B) [ – sin x = sin (– x)]
= 2sinC D C D
sin2 2
+ − −
Using 2
= 2 sinC D D C
sin2 2
+ −
cos C – cos D = 2 sinC D D C
sin2 2
+ −
Transformation of a Product into a Sum or
Difference 7
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Express the following as a sum or difference:
1. 2 sin 5x cos 3x
Sol. 2 sin 5x cos 3x = sin (5x + 3x) + sin (5x – 3x)
= sin 8x + sin2x
2. 7x 5x
2cos sin2 2
Sol. 7x 5x
2cos sin2 2
= + − −
7x 5x 7x 5xsin sin
2 2 2 2
[ 2 cos A sin B = sin (A + B) – sin (A – B)]
= sin 6x – sinx
3. x 3x
sin sin2 2
Sol. x 3x 1 x 3xsin sin 2sin sin
2 2 2 2 2
=
= 1 x 3x x 3x
cos cos2 2 2 2 2
− − +
[ 2 sin A sin B = cos (A – B) – cos (A + B)
= 1
2 [cos (–x) – cos 2x] =
1
2 [cos x – cos 2x]
[ cos (– x) = cos x]
Ex.4. cosec (45° + x) cosec (45° – x) =?
Sol. cosec (45° + x) cosec (45° – x)
= ( ) ( )
2
2sin 45 x sin 45 x + −
= ( ) ( ) ( ) ( )
2
cos 45 x 45 x cos 45 x 45 x + − − − + + −
[ 2 sin A sin B = cos (A – B) – cos (A + B)]
= 2 2
cos2x cos90 cos2x 0=
− − = 2 sec 2x
5. cos 5x – cos 11x =?
Sol. cos 5x – cos 11x
= 2 sin 5x 11x 11x 5x
sin2 2
+ −
C D D CcosC cosD 2sin sin
2 2
+ − − =
= 2 sin 8x sin 3x
6. sin 80° - sin 20°=?
Sol. sin 80° - sin 20°
= 2cos 80 20 80 20
sin2 2
+ −
C D C DsinC sinD 2cos sin
2 2
+ − − =
= 2 cos 50° sin 30° = 2 cos 50°1
2 = cos 50°
7. sin3A sin5A
cos3A cos5A
+
+ =?
Sol. sin3A sin5A
cos3A cos5A
+
+
=
+ −
+ −
5A 3A 5A 3A2sin cos
2 2
5A 3A 5A 3A2cos cos
2 2
= 2sin4AcosA
2cos4AcosA= tan4A
8. sin 65° + cos 65° = ?
Sol. sin 65° + cos 65° = sin 65° + cos (90° – 25°)
= sin 65° + sin 25°
= 65 25 65 252sin cos
2 2
+ −
= 2 sin 45°cos 20° = 2 × 1
2 × cos 20° = 2 cos20
9. +
+
sinA sinB
cosA cosB=?
Sol. +
+
sinA sinB
cosA cosB=
+ −
+ −
A B A B2sin cos
2 2A B A B
2cos cos2 2
= A B
tan2
+
10. cot x (sin 5x – sin 3x)
Sol. cot x (sin 5x – sin 3x)
= cot x 5x 3x 5x 3x
2cos sin2 2
+ −
A B A BsinA sinB 2cos sin
2 2
+ − − =
Examples
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= cosx
sinx × 2 cos 4x. sin x = 2 cos 4x cos x
10. sin 10° + sin 20° + sin 40° + sin 50° = ?
Sol. (sin 50° + sin 10°) + (sin 40° + sin 20°)
= 50 10 50 102sin cos
2 2
+ −
40 20 40 202sin cos
2 2
+ − +
= 2 sin 30° cos 20° + 2 sin 30° cos 10°
= 2 × 1
2 × cos 20° + 2 ×
1
2 × cos 10°
= cos 20° + cos 10°
11. cos 20° cos 40° cos 60° cos 80° =?
Sol. cos 60° cos 80° cos 40° cos 20°
= 1
2 cos 80° cos 40° cos 20°
1cos60
2
=
= 1
4 (2 cos 80° cos 40°) cos 20°
= 1
4 [cos (80° + 40°) + cos (80° – 40°)] cos 20°
= 1
4 [cos 120° + cos 40°] cos 20°
= 1 1cos40 cos20
4 2
− +
= 1 1cos20 cos40 cos20
4 2
− +
= 1
8− cos 20° +
1
8 [2 cos 40° cos 20°]
= 1
8− cos 20° +
1
8 [cos (40° + 20°) + cos (40° – 20°)]
= 1
8− cos 20° +
1
8 [cos 60° + cos 20°]
= 1 1 1cos20 cos20
8 8 2
− + +
= 1
8− cos 20° +
1 1 1cos20
16 8 16+ =
12. 4 cos 12° cos 48° cos 72° =?
Sol. L.H.S. = 4 cos 12° cos 48° cos 72°
= 2 (2 cos 48° cos 12°) cos 72°
= 2 [cos (48° + 12°) + cos (48° – 12°)] cos 72°
[ 2 cos A cos B = cos (A + B) + cos (A – B)]
= 2 (cos 60° + cos 36°) cos 72°
= 2 cos 60° cos 72° + 2 cos 36° cos 72°
= 2 × 1
2 cos 72° + 2 cos 72° cos 36°
= cos 72° + [cos (72° + 36°) + cos (72° – 36°)]
= cos 72° + cos 108° + cos 36°
= cos 72° + cos (108° – 72°) + cos 36°
= cos 72° – cos 72° + cos 36°
[ cos (180° – x) = – cos x] = cos 36°
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1. 5
cos cos12 12
=?
(a) 1 (b) 1
4 (c) 1/2 (d)
3
2
2. 2 sin 45° sin 15° =?
(a) 1 (b) 3 1
2
+ (c) 1/2 (d)
3 1
2
−
3. sin 14x + sin 2x =?
(a) 2 sin 8x cos 6x (b) 2 sin 6x cos 8x
(c) 1/2 (d) sin 4x cos 10x
4. 1
2 (cos 7x + cos x)=?
(a) cos 4x cos 3x (b) sin 4x sin 3x
(c) cos 4x sin 3x (d) 1
5. sin 47° + cos 77° = ?
(a) sin 17° (b) cos 17°
(c) 2 sin 73° (d) 2cos73°
6. sin 25° cos 115° = ?
(a) 1
2 (cos 40° – 1) (b)
1
2 (cos 50° – 1)
(c) 1
2 (sin 40° – 1) (d) 1
7. cos 80° + cos 40° – cos 20° = ?
(a) 1 (b) 0 (c) 1/2 (d) 2
8. sin50 sin70 sin10− + =?
(a) 1 (b) 1/2 (c) 0 (d) 2
9. sin7A sin5A
cos7A cos5A
−
+ =?
(a) cot 6A (b) cot A (c) tan 6A (d) tan A
10. −
−
cos9x cos5x
sin17x sin3x=?
(a) sin2x
cos10x
− (b)
cos2x
sin10x
− (c) tan 7x (d) cot 7x
11. cot 4x (sin 5x + sin 3x)=?
(a) 2 sin 4x cos x (b) 2 cos 4x sin x
(c) 2 cos 4x cos x (d) 2
12. The value of ++ 172cos68cos52cos is
(a) 0 (b) 1 (c) 2 (d) 2
3
13. cos 55° + cos 65° + cos 175° = ?
(a) 1 (b) 0 (c) 1/2 (d) 2
14. cos7A cos11A
cos7A cos11A
+
− =?
(a) 1 (b) cot 9
tan 2A
A (c)
tan9
tan 2A
A (d)
tan9
cot 2A
A
15. cos 8 – cos 3 =
(a)
−11 5
2sin sin2 2
(b) 13 3
2sin sin2 2
−
(c) 11 5
2cos cos2 2
− (d)
11 52sin cos
2 2
16. 5 2
cos sin7 7
=?
(a) 1 3
cos2 7
(b) 0 (c) -
1 3sin
2 7
(d)-1
17. cos 35°cos 25°=?
(a) 1
sin102
(b) 1
cos102
(c) 1/2 (d) 1 1
cos104 2
+
18. cos 15° – sin 15° = ?
(a) 3
2 (b) 3
2 2 (c) 1 (d)
1
2
19. If ,)sin(
)sin(
ba
ba
yx
yx
−
+=
−
+then
y
x
tan
tan is equal to
(a) a
b (b) b
a (c) ab (d) None of these
20. If cos (A – B) = 3 cos (A + B) then find the value of tan A
tan B
Exercise - 1
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(a) 1 (b) 2 (c) 1
2 (d)
1
2−
21. =+++
+++
9cos7cos5cos3cos
9sin7sin5sin3sin
(a) 3tan (b) 3cot (c) 6tan (d) 6cot
22. 9 3 5
2cos cos cos cos13 13 13 13
+ + =?
(a) – 1 (b) 0 (c) 1 (d) None of these
23. If ,cotcoscos
sinsinB
AC
CA=
−
− then A,B,C are in
(a) A.P. (b) G.P.
(c) H.P. (d) None of these
24. If ,tancostan = then =2
tan2
(a) )sin(
)sin(
−
+ (b)
)cos(
)cos(
+
−
(c) )sin(
)sin(
+
− (d)
)cos(
)cos(
−
+
25. If ,cos21
=+x
x then =+3
3 1
xx
(a) 3cos (b) 3cos2
(c) 3cos2
1 (d) 3cos3
1
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1. cot 4x cos3x cos2x
sin4x sin3x sin2x
+ +
+ + =?
(a) cot 3x (b) cot2x (c) tan3x (d) tan2x
2. sin 10° + sin 20° + sin 40° + sin 50° =?
(a) 2 sin 15° sin 5° (b) 2 sin 15° cos 5°
(c) 2 cos 15° sin 5° (d) 2 cos 15° cos 5°
3. =++−
−++
)cos()sin(
)cos()sin(
ABAB
ABAB
(a) BB
BB
sincos
sincos
−
+ (b) AA
AA
sincos
sincos
−
+
(c) AA
AA
sincos
sincos
+
− (d) None of these
4. 9 3 5
2cos cos cos cos13 13 13 13
+ + =?
(a) 3 (b) 2 (c) 0 (d) 1
5. cos2 A + cos2 B – 2 cos A cos B cos (A + B) =?
(a) sin2 (A + B) (b) cos2 (A+B)
(c) sin2 (A - B) (d) cos2 (A - B)
6. ( ) ( )( ) ( ) − − −
− − −
2sin cos sin 2
2sin cos sin 2=?
(a)
sin
sin (b)
cos
cos (c)
cos
cos (d)
sin
sin
7. cos 7x + cos 5x + cos 3x + cos x =?
(a) 4 cos x cos 2x cos 4x
(b) 2 cos x cos 2x cos 4x
(c) cos x cos 2x cos 4x
(d) 4 cos x cos 3x cos 5x
8. If cosec A + sec A = cosec B + Sec B then find the value of
cotA B
2
+
(a) tan A tan B (b) cot A cot B
(c) tan A cot B (d) cot A tan B
9. If three angles A, B and C are in A.P. then find the value
of sin A sinC
cosC cosA
−
−.
(a) cot (A + C) (b) cot B
(c) cot A (d) tan (B + C)
10. If ),2sin(sin += ab then =−
+
ba
ba
(a) )tan(
tan
+ (b)
)cot(
cot
−
(c) )cot(
cot
+
− (d)
)cot(
cot
+
11. cos 2x cos x
2 – cos 3x cos
9x
2 =?
(a) 5x
sin5xsin2
(b) cos 5x cos5
2
x
(c) sin 5x cos5
2
x (d) cos 5x sin
5
2
x
12. If sin3sin2sinsin =++
and cos3cos2coscos =++ , then is equal to
(a) 2/ (b) (c) 2 (d) 6/
13. =−− xxx 5cos3coscos2
(a) xx 23 sincos16 (b) xx 23 cossin16
(c) xx 23 sincos4 (d) xx 23 cossin4
14. If ,coscos BmA = then
(a) 2
tan1
1
2cot
AB
m
mBA −
−
+=
+
(b) 2
cot1
1
2tan
AB
m
mBA −
−
+=
+
(c) 2
tan1
1
2cot
BA
m
mBA −
−
+=
+
(d) None of these
15. If sin A = sin B and cos A = cos B then which of the
following is true
(a) A Bcos
2
+
= 0 (b) A Bsin
2
−
= 0
(c) A Bsin
2
+
= 0 (d) None is true
16. If 0coscoscos =++ yx and ,0sinsinsin =++ yx then
Exercise - 2
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=
+
2cot
yx?
(a) sin (b) cos
(c) cot (d)
+
2sin
yx
17. The value of oo 70cos470cot + is
(a) 3
1 (b) 3 (c) 32 (d)
2
1
18. sin8xcosx sin6xcos3x
cos2xcosx sin3xsin4x
−
− =?
(a) tan3x (b) cot2x (c) sec2x (d) tan 2x
19. cos6x 6cos4x 15cos2x 10
cos5x 5cos3x 10cosx
+ + +
+ + =?
(a) sin2x (b) cos3x (c) 3sin2x (d) 2cos x
20. If ,coscos,sinsin DBACBA =+=+ then the value of
=+ )sin( BA
(a) CD (b) 22 DC
CD
+
(c) CD
DC
2
22 + (d) 22
2
DC
CD
+
21. =+−− 81tan63tan27tan9tan
(a) 1/2 (b) 2 (c) 4 (d) 8
22. If tan1 = k cot 2 then ( )( )
1 2
1 2
cos
cos
−
+ =
(a) 1 k
1 k
+
− (b)
1 k
1 k
−
− (c)
k 1
k 1
+
− (d)
k 1
k 1
−
+
23. Find the value
+ +
+ +
cos3 2cos5 cos7
cos 2cos3 cos5 + sin 2 tan 3
(a) cos 2 (b) tan 2 (c) cos 4 (d) sec 2
24. + + +
+ + +
sin sin2 sin4 sin5
cos cos2 cos4 cos5
A A A A
A A A A =?
(a) tan8A (b) cotA (c) tan6A (d) tan3A
25. In quadrilateral ABCD if sin 2
A B+
cos2
A B−
sin cos 22 2
C D C D+ − + =
, then value of sin
2
Asin
2
Bsin
2
Csin
2
D
(a) 0 (b) 1
16 (c)
1
4 (d) 1
26. 2 sin (60° – x) cos (30° + x)
(a) 1 - +cos2x 3 sin2x
2 2 (b) 1 -
cos2x 3 sin2x
2 2−
(c) 0 (d) 1 + cos2x 3 sin2x
2 2−
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Answer key
1 b 2 d 3 a 4 a 5 b
6 c 7 b 8 c 9 d 10 a
11 c 12 a 13 b 14 b 15 a
16 c 17 d 18 b 19 c 20 c
21 c 22 b 23 a 24 c 25 b
1. (b)
5 1 5cos cos 2cos cos
12 12 2 12 12
=
= 1 5 5
cos cos2 12 12 12 12
+ + −
= 1
cos cos2 2 3
+
= 1 1
02 2
+
=
1
4
2. (d)
2 sin 45° sin 15°
= cos (45° – 15°) – cos (45° + 15°)
= cos 30° - cos 60° = 3 1 3 1
2 2 2
−− =
3. (a)
sin 14x + sin 2x
= 2 sin14x 2x 14x 2x
cos2 2
+ −
C D C DsinC sinD 2sin cos
2 2
+ − + =
= 2 sin 8x cos 6x
4. (a)
1
2 (cos 7x + cos x) = 1 7x x 7x x
2cos cos2 2 2
+ −
C D C DcosC cosD 2cos cos
2 2
+ − + =
= cos 4x cos 3x
5. (b)
sin 47° + cos 77°
= sin 47° + cos 13° = 2 sin 30°cos 17° = cos 17°
6. (c)
sin 25° cos 115°
= 1
2 (2 sin 25° cos 115°) =
1
2 (sin 140° – sin 90°)
= 1
2 [sin (180° – 40°) – 1] =
1
2 (sin 40° – 1)
7. (b)
cos 80° + cos 40° – cos 20°
= 2 cos 60°.cos 20° – cos 20°
= cos 20° – cos 20° = 0
8. (c)
ooo 10sin70sin50sin +−
2cos60 sin10 sin10o o o= − + sin10 (1 2 cos 60 ) 0o o= − =
9. (d)
sin7A sin5A
cos7A cos5A
−
+ =
A 5A A 5A2sin cos
2 2
A 5A A 5A2cos cos
2 2
− +
+ −
= 2sinAcos6A
2cos6AcosA = tan A
10. (a)
−
−
cos9x cos5x
sin17x sin3x
=
+ − −
+ −
9x 5x 9x 5x2sin sin
2 2
17x 3x 17x 3x2cos sin
2 2
= 2sin7xsin2x sin2x
2cos10xsin7x cos10x
− −=
11. (c)
cot 4x (sin 5x + sin 3x)
Solutions
Exercise - 1
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= cot 4x 5x 3x 5x 3x
2sin cos2 2
+ −
A B A BsinA sinB 2sin cos
2 2
+ − + =
= cos4x
sin4x × 2 sin 4x cos x = 2 cos 4x cos x
12. (a)
ooo 172cos68cos52cos ++
ooo 68cos)172cos52(cos ++=
ooo 68cos60cos112cos2 +=
cos 112 cos 68o o= + 2 cos (90 ) cos 22 0o o= =
13. (b)
cos 55° + cos 65° + cos 175°
= 55 65 65 552cos cos cos175
2 2
+ − +
= 2 cos 60°.cos 5° + cos 175°
= 2 × 1
2 × cos 5° + cos (180° – 5°) = cos 5° – cos 5° = 0
14. (b)
cos7A cos11A
cos7A cos11A
+
−=
+ −
+ −
7A 11A 11A 7A2cos cos
2 2
7A 11A 11A 7A2sin sin
2 2
= cos9Acos2A
sin9Asin2A = cot 9A cot 2A =
cot 9
tan 2A
A
15. (a)
cos 8 – cos 3
= + −
8 3 3 82sin sin
2 2
= −
11 52sin sin
2 2 =
−
11 52sin sin
2 2
16. (c)
5 2
cos sin7 7
=
1 5 22cos sin
2 7 7
= 1 5 2 5 2
sin sin2 7 7 7 7
+ − −
= 1 3sin sin
2 7
−
= -1 3
sin2 7
17. (d)
1
2 (cos 35°cos 25°)
= 1
2 [cos (35° + 25°) + cos (35° – 25°)]
= 1
2 (cos 60° + cos 10°)
= +
1 1cos10
2 2 =
1 1cos10
4 2+
18. (d)
cos 15° – sin 15° = sin 75° – sin 15°
= 2cos75 15
2
+ sin
75 15
2
−
= 2cos 45° sin 30° = =1 1 1
222 2
19. (b)
ba
ba
yx
yx
−
+=
−
+
)(sin
)(sin
)()(
)()(
)(sin)(sin
)(sin)(sin
baba
baba
yxyx
yxyx
−−+
−++=
−−+
−++
2sin cos 2
2cos sin 2
x y a
x y b =
tan
tan
x a
y b =
20. (c)
( )( )
cos A B
cos A B
−
+ = 3
Applying C & D
( ) ( )( ) ( )
cos A B cos A B 3 1
3 1cos A B cos A B
− + + +=
−− − +
2cosAcosB
2sinAsinB = 2 tan A tan B =
1
2
21. (c)
9cos7cos5cos3cos
9sin7sin5sin3sin
+++
+++
)7cos5(cos)9cos3(cos
)7sin5(sin)9sin3(sin
+++
+++=
cos6cos23cos6cos2
cos6sin23cos6sin2
+
+=
2sin 6 (cos3 cos )
2cos6 (cos3 cos )
+=
+
= tan6
22. (b)
9 3 5
2cos cos cos cos13 13 13 13
+ +
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9 4
2cos cos 2cos cos13 13 13 13
= +
+=
13
4cos
13
9cos
13cos2
5
2cos 2cos cos 013 2 26
= =
,
= 0
2cos
23. (a)
BAC
CAcot
coscos
sinsin=
−
− BCACA
CACA
cot
2sin
2sin2
2sin
2cos2
=−+
−+
BCA
cot2
)(cot =
+
2
CAB
+=
Thus A, B, C will be in A.P.
24. (c)
2 1 costan
2 1 cos
−=
+ =
tan tan
tan tan
−
+=
sin( )
sin( )
−
+
25. (b)
cos21
=+x
x
+−
+=+
xx
xx
xx
xx
113
113
3
3
= 3(2cos ) 3(2cos ) − = 38cos 6cos −
= 32(4cos 3cos ) − = 2cos3
Alternate:
Put 1=x = 0
Then 3cos2213
3 ==+x
x
Answer key
1 a 2 d 3 b 4 c 5 a
6 d 7 a 8 a 9 b 10 c
11 c 12 a 13 a 14 a 15 b
16 c 17 b 18 d 19 d 20 d
21 c 22 a 23 a 24 d 25 c
26 d
1. (a)
cot 4x cos3x cos2x
sin4x sin3x sin2x
+ +
+ +
= ( )( )
cos4x cos2x cos3x
sin4x sin2x sin3x
+ +
+ +
=
6x 2x2cos .cos cos3x
2 2
6x 2x2cos .cos sin3x
2 2
+
+
= 2cos3xcosx cos3x
2sin3xcosx sin3x
+
+ =
( )( )
cos3x 2cosx 1
sin3x 2cosx 1
+
+ = cot 3x
2. (d)
sin 10° + sin 20° + sin 40° + sin 50°
= (sin 50° + sin 10°) + (sin 40° + sin 20°)
= 50 10 50 102sin cos
2 2
+ −
40 20 40 202sin cos
2 2
+ − +
= 2 sin 30° cos 20° + 2 sin 30° cos 10°
= 2 sin 30° [cos 20° + cos 10°]
= 2 sin 30°20 10 20 10
2cos cos2 2
+ −
= 2 × 1
2 [2 cos 15° cos 5°] = 2 cos 15° cos 5°
3. (b)
)(cos)(sin
)(cos)(sin
ABAB
ABAB
++−
−++
( )( )
sin ( ) sin[90 ]
sin ( ) sin[90 ]
o
o
B A B A
B A A B
+ + − −=
− + − −
)45(cos)45(sin2
)45(cos)45(sin2
BA
BAoo
oo
−−
−+=
Exercise - 2
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sin ( 45 )
sin (45 )
o
o
A
A
+=
− =
cos sin
cos sin
A A
A A
+
−
4. (c)
9 3 5
2cos cos cos cos13 13 13 13
+ +
= 9 9 3 5cos cos cos cos
13 13 13 13 13 13
+ + − + +
[ 2 cos A cos B = cos (A + B) + cos (A – B)]
= 10 8 3 5
cos cos cos cos13 13 13 13
+ + +
= 3 5 3 5cos cos cos cos
13 13 13 13
− + − + +
= 3 5 3 5
cos cos cos cos13 13 13 13
− − + +
[ cos ( – x) = – cos x] = 0
5. (a)
cos2 A + cos2 B – 2 cos A cos B cos (A + B)
= cos2 A + cos2 B – [cos (A + B) + cos (A – B)] cos (A + B)
= cos2 A + cos2 B – cos2 (A + B) – cos (A – B) cos (A + B)
= cos2 A + cos2 B – cos2 (A + B) – (cos2 A – sin2 B)
= (sin2 B + cos2 B) – cos2 (A + B)
= 1 – cos2 (A + B) = sin2 (A + B)
6. (d)
( ) ( )( ) ( )
2sin cos sin 2
2sin cos sin 2
− − −
− − −
= ( ) ( ) ( )( ) ( ) ( )
sin sin sin 2
sin sin sin 2
− + + − − − −
− + + − − − −
[ 2 sin A cos B = sin (A + B) + sin (A – B)]
= ( ) ( )( ) ( )
sin sin 2 sin 2
sin sin 2 sin 2
+ − − −
+ − − − =
sin
sin
7. (a)
cos 7x + cos 5x + cos 3x + cos x
= (cos 7x + cos 3x) + (cos 5x + cos x)
= 7x 3x 7x 3x
2cos cos2 2
+ −
5x x 5x x
2cos cos2 2
+ − +
= (2 cos 5x cos 2x) + (2 cos 3x cos 2x)
= 2 cos 2x [cos 5x + cos 3x]
= 2 cos 5x 3x 5x 3x
2x 2cos cos2 2
+ −
= 2 cos 2x [2 cos 4x. cos x] = 4 cos x cos 2x cos 4x
8. (a)
cosec A + sec A = cosec B + sec B
cosec A – cosec B = sec B – sec A
1 1 1 1
sin sin cosB cosAA B− = −
− −
=sinB sinA cosA cosB
sinAsinB cosAcosB
+ − + −
=
A B B A A B B A2cos sin 2sin sin
2 2 2 2
sinAsinB cosAcosB
+
=+
A Bcos
sin AsinB2
A B cos AcosBsin
2
cotA B
2
+ = tan A tan B
9. (b)
A, B & C are in A.P.
So, B = A C
2
+
sin A sinc
cosC cosA
−
− =
+ −
+ −
A C A C2cos sin
2 2
A C A C2sin sin
2 2
= cot +
A C
2= cot B
10. (c)
sin sin ( 2 )b a = +
sin
sin ( 2 )
a
b
=
+
sin sin ( 2 )
sin sin ( 2 )
a b
a b
+ + + =
− − +
2sin ( )cos
2cos ( )sin
+=
− + tan ( )cot = − +
cot
cot ( )
= −
+
11. (c)
cos 2x cos x
2 – cos 3x cos
9x
2
= 1 x 9x2cos2xcos 2cos cos3x
2 2 2
−
= 1 x x
cos 2x cos 2x2 2 2
+ + −
9x 9xcos 3x cos 3x
2 2
− + + −
= 1 5x 3x 5x 3xcos cos cos cos
2 2 2 2 2
+ − −
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= 1 5x 15xcos cos
2 2 2
−
=
+ −
5x 15x 15x 5x1 2 2 2 22sin sin2 2 2
= 10x 5x
sin sin2 2
= 5x
sin5xsin2
12. (a)
sin2sin3sinsin =++
sin2sincos2sin2 =+
sin)1cos2(2sin =+ ….. (i)
Now cos2cos3coscos =++
cos2coscos2cos2 =+
cos)1cos2(2cos =+ ….. (ii)
Divide (i) by (ii)
tan2tan =
=2 2/ =
13. (a)
2cos cos3 cos5x x x− − ( )2cos cos3 cos5x x x= − +
2cos 2cos cos4x x x= − 2cos (1 cos4 )x x= −
xx 2sin2cos2 2=24cos sin 2x x= 2 316sin cosx x=
14. (a)
cos cosA m B= cos
1 cos
m A
B=
1 cos cos
1 cos cos
m A B
m A B
+ + =
− −=
2cos cos2 2
2sin sin2 2
A B B A
A B B A
+ −
+ −
−
+=
2cot
2cot
ABBA
Hence, 2
tan1
1
2cot
AB
m
mBA −
−
+=
+
15. (b)
sin A = sin B
sin A – sin B = 0
A B A B2sin cos
2 2
− +
= 0 ….. (i)
cos A = cos B
cos A – cos B = 0
A B A B2sin cos
2 2
− + −
= 0 ….. (ii)
From these 2 equations we can conclude that
A Bsin
2
−
= 0 ( it is in both the equations)
16. (c)
0coscoscos =++ yx coscoscos −=+ yx
cos2
cos2
cos2 −=
−
+ yxyx …..(i)
and sin sin sin 0x y + + = .sinsinsin −=+ yx
sin2
cos2
sin2 −=
−
+ yxyx ….. (ii)
Divide (i) by (ii)
−
+
−
+
2cos
2sin2
2cos
2cos2
yxyx
yxyx
sin
cos= cot
2cot =
+ yx
17. (b)
o
ooooo
70sin
70cos70sin470cos70cos470cot
+=+
cos70 2sin140
sin 70
o o
o
+=
cos70 2sin(180 40 )
sin70
o o o
o
+ −=
sin 20 sin 40 sin 40
sin70
o o o
o
+ +=
2sin30 cos10 sin 40
sin 70
o o o
o
+=
sin80 sin 40
sin70
o o
o
+=
2sin60 cos203
sin70
o o
o= =
18. (d)
sin8xcosx sin6xcos3x
cos2xcosx sin3xsin4x
−
−
=
12sin8xcosx 2sin6xcos3x
21
2cos2xcosx 2sin3xsin4x2
−
−
= ( ) ( ) ( ) ( )
( ) ( ) ( ) ( )
sin 8x x sin 8x x sin 6x 3x sin 6x 3x
cos 2x x cos 2x x cos 3x 4x cos 3x 4x
+ + − − + + −
+ + − − + + −
= sin9x sin7x sin9x sin3x
cos3x cosx cosx cos7x
+ − +
+ − +
[ cos (– x) = cos x]
= −
+
sin7x sin3x
cos3x cos7x=
+ −
+ −
7x 3x 7x 3x2cos sin
2 2
3x 7x 3x 7x2cos cos
2 2
= ( )
cos5xsin2x sin2x
cos2xcos5xcos 2x=
−
[ cos (– 2x) = cos 2x] = tan 2x
19. (d)
cos 6x + 6 cos 4x + 15 cos 2x + 10
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= (cos 6x + cos 4x) + (5 cos 4x + 5 cos 2x) + (10 cos 2x +
10)
= (cos 6x + cos 4x) + 5 (cos 4x + cos 2x + 10 (cos 2x +
cos 0x)
=
10x 2x2cos cos
2 2+
6x 2x5 2cos cos
2 2+
2x 2x10 2cos .cos
2 2
= 2 cos 5x cos x + 5 (2 cos 3x cos x) + 10 (2 cos x cos x)
= 2 cos x [cos 5x + 5 cos 3x + 10 cos x]
Now,
cos6x 6cos4x 15cos2x 10
cos5x 5cos3x 10cosx
+ + +
+ +
= ( )2cosx cos5x 5cos3x 10cosx
cos5x 5cos3x 10cosx
+ +
+ + = 2 cos x
20. (d)
D
C
BA
BA=
+
+
coscos
sinsin
2 sin cos2 2
2cos cos2 2
A B A B
C
A B A B D
+ −
=+ −
tan2
A B C
D
+ =
Now, 2
2 tan2
sin ( )
1 tan2
A B
A BA B
+
+ =+
+
=2 2 2
2
22
( )1
C
CDD
C C D
D
=+
+
21. (c)
oooo 81tan63tan27tan9tan +−−
oooo 9cot27cot27tan9tan +−−=
)27cot27(tan)9cot9(tan oooo +−+=
= sin9 cos9 sin 27 cos27
cos9 sin9 cos27 sin 27
+ − +
cos(9 9 ) cos(27 27 )
sin9 cos9 sin 27 cos27
o o o o
o o o o
− −= −
2 2
sin18 sin54o o= − sin 54 sin 18
2sin 18 sin 54
o o
o o
−=
2cos36 sin18
2 4sin18 sin54
o o
o o= =
22. (a)
Given that tan 1 tan 2 = k
( )( )
1 2
1 2
cos
cos
−
+ = 1 2 1 2
1 2 1 2
cos cos sin sin
cos cos sin sin
+
−
= 1 2
1 2
1 tan tan 1 k
1 tan tan 1 k
+ +=
− −
23. (a)
+ +
+ +
cos3 2cos5 cos7
cos 2cos3 cos5 + sin 2 tan 3
= + +
+ +
cos3 cos7 2cos5
cos cos5 2cos3+ sin 2 tan 3
= 2cos5 .cos2 2cos5
2cos3 cos2 2cos3
+
+ + sin 2 tan 3
=
+
+
2cos5 cos2 1
2cos3 cos2 1 + sin 2 tan 3
= cos5 sin2 sin3
cos3 cos3
+
[cos5 = cos (2 + 3)]
= cos2 cos3 sin2 sin3 sin2 sin3
cos3
− +
= cos 2
24. (d)
sin sin2 sin4 sin5
cos cos2 cos4 cos5
A A A A
A A A A
+ + +
+ + +
= ( ) ( )( ) ( )
sin5 sin sin4 sin2
cos5 cos cos4 cos2
A A A A
A A A A
+ + +
+ + +
2sin3 cos2 2sin3 cos
2cos3 cos2 2cos3 cos
A A A A
A A A A
+=
+
( )( )
2sin3 cos2 cos
2cos3 cos2 cos
A A A
A A A
+=
+= tan3A
25. (c)
sin cos sin cos2 2 2 2
A B A B C D C D+ − + − +
=2
1
2[sinA + sinB+ sin C + sinD] =2
sin A + sin B + sinC + sin D = 4
A = B = C = D = 90°
So, sin1
sin sin sin2 2 2 2 4
A B C D=
26. (d)
2 sin (60° – x) cos (30° + x)
= sin (60° – x + 30° + x) + sin (60° – x – 30° – x)
= sin 90° + sin (30 – 2x)
= 1 + sin 30° cos 2x – cos 30° sin 2x
= 1 + cos2x 3 sin2x
2 2−
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1. Trigonometric functions of 2A in terms of A.
(i) sin 2A = 2 sin A cos A
(ii) cos 2A = cos2 A – sin2 A
= 2 cos2 A – 1
= 1 – 2 sin2 A
(iii) tan 2A = 2
2tan A
1 tan A−
(iv) sin 2A = 2
2tan A
1 tan A+
(v) cos 2A = 2
2
1 tan A
1 tan A
−
+
(i) sin 2A = sin (A + A)
= sin A cos A + cos A sin A
= 2 sin A cos A
sin 2A = 2 sin A cos A
(ii) cos 2A = cos (A + A)
= cos A cos A – sin A sin A
= cos2 A – sin2 A
cos 2A = cos2 A – sin2 A
= cos2 A – (1 – cos2 A)
= 2 cos2 A – 1
cos 2A = cos2 A – sin2 A
= (1 – sin2 A) – sin2 A
= 1 – 2 sin2 A
(iii) tan 2A = tan (A + A)
= 2
tanA tanA 2tanA
1 tanAtanA 1 tan A
+=
− −
tan 2A = 2
2tan A
1 tan A−
(iv) sin 2A = 2sinAcosA
1
[ cos2 A + sin2 A = 1]
= +2 2
2sinAcosA
cos A sin A
= 2
2 2
2
2sin Acos A
cos Acos A sin A
cos A
+
= 2
2tan A
1 tan A+
(v) cos 2A = −
+
2 2
2 2
cos A sin A
cos A sin A
=
−
+
2 2
2
2 2
2
cos A sin A
cos Acos A sin A
cos A
= 2
2
1 tan A
1 tan A
−
+
------------------* * *------------------
(1) cos 2A = 1 – 2 sin2 A
2 sin2 A = 1 – cos 2A
sin2 A = 1– cos2
2
A
(2) cos 2A = 2 cos2 A – 1
2 cos2 A = 1 + cos 2A
cos2 A = 1 cos2
2
A+
2. Trigonometric functions of 3A in terms of A
(i) sin 3A = 3 sin A – 4 sin3 A
(ii) cos 3A = 4 cos3 A – 3 cos A
(iii) tan 3A = 3
2
3tanA tan A
1 3tan A
−
−
(i) sin 3A = sin (A + 2A)
= sin A cos 2A + cos A sin 2A
= sin A(1 – 2 sin2 A) + cos A 2 sin A cos A
[ cos 2A = 1 – 2 sin2 A and sin 2A = 2 sin A cos A]
= sin A – 2 sin3 A + 2 sin A (1 – sin2 A)
= 3 sin A – 4 sin3 A
Multiples and sub Multiples 8
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sin 3A = 3 sin A – 4 sin3 A
(ii) cos 3A = cos (A + 2A)
= cos A cos 2A – sin A sin 2A
= cos A (2 cos2 A – 1) – sin A 2 sin A cos A
= 2 cos3 A – cos A – 2 cos A sin2 A
= 2 cos3 A – cos A – 2 cos A (1 – cos2 A)
= 4 cos3 A – 3 cos A
cos 3A = 4 cos3 A – 3 cos A
(iii) tan 3A = tan (A + 2A)
= tanA tan2A
1 tanAtan2A
+
−
= +
−
−−
2
2
2tan Atan A
1 tan A2tan A
1 tan A1 tan A
=− +
− −
3
2 2
tanA tan A 2tanA
1 tan A 2tan A
= 3
2
3tanA tan A
1 3tan A
−
−
tan 3A = 3
2
3tan A tan A
1 3tan A
−
−
3. Trigonometric functions of A in terms of A
2.
Replacing A by A
2 in the results of trigonometric
functions of 2A in terms of those of A, we get
(i) sin A = 2 sinA A
cos2 2
(ii) cos A = 2 2A A
cos sin2 2−
= 1 – 2 sin2A
2
= 2 cos2 A
2 – 1
(iii) 2 A 1 cosA
sin2 2
−=
(iv) 2 A 1 cosA
cos2 2
+=
(v) tan A = 2
A2tan
2A
1 tan2
−
(vi) sin A = 2
A2tan
2A
1 tan2
+
(vii) cos A =
2
2
A1 tan
2A
1 tan2
−
+
TRIGONOMETRIC FUNCTIONS OF
SOME PARTICULAR ANGLES
Values at 18°, 36°
Let x = 18°. Then 5x = 90°
2x + 3x = 90° 2x = 90° – 3x
sin 2x = sin (90° – 3x) = cos 3x
2 sin x cos x = 4 cos3 x – 3 cos x
2 sin x = 4 cos2 x – 3
2 sin x = 4 (1 – sin2 x) – 3
2 sin x = 1 – 4 sin2 x
4 sin2 x + 2 sin x – 1 = 0
Which is a quadratic in sin x
sin x = 2 4 16 2 20
8 8
− + − =
= 2 2 5 1 5
8 4
− − =
sin 18° = 1 5
4
−
Since 18° lies in the first quadrant, sin 18° is +ve. Hence
rejecting the –ve sign we get
sin18° = 5 1
4
−
Also cos2 x = 1 – sin2 x
cos2 18° = 1 – sin2 18° =
2
5 11
4
−−
= + −
−5 1 2 5
116
= ( )16 5 1 2 5 10 2 5
16 16
− + − +=
cos 18° = 10 2 5 10 2 5
16 4
+ + =
[Taking +ve sign before the square root, since 18° is in
1st quadrant and in 1st quadrant cosine is +ve]
(i) sin 72° = sin (90° – 18°)
= cos 18° = 10 2 5
4
+
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(ii) cos 72° = cos (90° – 18°) = sin 18° = 5 1
4
−
(iii) cos 36° = 1 – 2 sin2 18° [ cos x = 1 – 2 sin2 x/2]
= 1 – 2
2
5 1
4
−
5 1
sin184
− =
= 1 – 2 + −
5 1 2 5
16 =
− + +=
8 6 2 5 5 1
8 4
(iv) sin 36° = − 21 cos 36 = +
−
2
5 11
4
= + +−
5 1 2 51
16=
−10 2 5
4 =
−10 2 5
4
(v) sin 54° = sin (90° – 36°) = cos 36° = 5 1
4
+
(vi) cos 54° = cos (90° – 36°) = sin 36° = 10 2 5
4
−
Values at 22.5° and 67.5°
tan 22.5° and cot 67.5°
tan2 = −
+
1 cos2
1 cos2
tan2 (22.5°)=1 cos45
1 cos45
−
+ =
12
12
−
+
= ( )22 1
2 12 1
−= −
+
tan 22.5° = 2 1−
cot 67.5° = cot (90° – 22.5°) = tan 22.5° = 2 1−
cot 22.5° and tan 67.5°
cot 22.5° = 1
tan22.5 =
12 1
2 1= +
−
tan (67.5°) = cot 22.5° = 2 1+
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90
1. (sinA + cosA)2 =?
Sol. (sin A + cosA)2
= sin2A +cos2A + 2 sinA cosA = 1 + sin2A
2. tan A + cot A =?
Sol. tan A + cot A
= sin cos
cos sin
A A
A A+ =
2 2sin cos
sin cos
A A
A A
+
= 1
sin cosA A=
2
2sin cosA A =
2
sin2A = 2cosec2A
3. −
+
1 cos2
1 cos2
A
A =?
Sol. 1 cos2
1 cos2
A
A
−
+ =
( )− −
+ −
2
2
1 1 2sin
1 2cos 1
A
A
= 2
2
1 1 2sin
2cos
A
A
− +=
2
2
2sin
2cos
A
A= tan2A
4. cos4 A – sin4 A =?
Sol. cos4A – sin4A= (cos2A)2 – (sin2A)2
= (cos2A + sin2A) (cos2A – sin2A) = cos 2 A
5. −
=+
1 cos?
1 cos
A
A
Sol. 1 cos
1 cos
A
A
−
+=
2
2
1 1 2sin2
1 2cos 12
A
A
− −
+ −
= 2
2sin cos2 2
1 1 2sin2
A A
A− +
= 2
2sin cos2 2
2sin2
A A
A=
cos2
sin2
A
A = cot
2
A
6. sin
1 cos
A
A−=?
Sol. sin
1 cos
A
A−=
2
2sin cos2 2
1 1 2sin2
A A
A − −
= 2
2sin cos2 2
2sin2
A A
A = cot
2
A
7. 1sin2cos2 22 =− , then =
Sol. 1sin2cos2 22 =− 12cos2 =
o60cos
2
12cos == oo 30602 ==
8. +
sin2
1 cos2
A
A =?
Sol. sin2
1 cos2
A
A+=
2
2sin cos
1 2cos 1
A A
A+ −=
2
2sin cos
2cos
A A
A=
sin
cos
A
A= tan A
9 Find value of sin2A if cosA = 3
2
Sol. cos A = 3
2
now, sin A = 21 cos A− =
2
31
2
−
= 3
14
− = 1
2
Now sin2A = 2sinAcosA = 2×1
2×
3
2 =
3
2
10. If tan A = 5
12then cos 2A=?
Sol. tanA = 5
12
now, sin 2A = 2
2tan
1 tan
A
A+ sin 2A =
2
52
12
51
12
+
=
5
625
1144
+
=
5
6169
144
= 144 5
6 169
=
120
169
Examples
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1. −
sin2
1 cos2
A
A =?
(a) cot A (b) tan A (c) cot 2A (d) tan 2A
2. cos4A – sin4 A =?
(a) sin 2A (b) 1 (c) cos 2A (d) 0
3. 2
2tan15
1 tan 15
− =?
(a) 3 (b) 1
3 (c) 1 (d) 0
4. (sinA - cosA)2 =?
(a) 1 + sin 2A (b) 1 – sin 2A
(c) sin A + cos A (d) tan 2A
5. sin2
1 cos2
A
A− =?
(a) tan A (b) tan 2A (c) cot A (d) cot 2A
6. cot A – tan A =?
(a) 2 tan 2A (b) 1 (c) sec 2A (d) 2 cot 2A
7. sin sin2
1 cos cos2
+
+ +=?
(a) tan θ (b) cot θ (c) 1 (d) tan 2θ
8. If cos A = 5
13 then sin 2A =?
(a) 12
13 (b)
24
13 (c)
120
169 (d)
13
24
9. If tan A = 1
3, then tan2A =?
(a) 2
3 (b) 3 (c) 2 3 (d) 1
10. If cos 2A = 13
36 then cos A =?
(a) 13
18 (b)
7
6 2 (c)
7
6 (d) 1
11. If sinA = 3
5then tan 2A =?
(a) 4
5 (b)
7
24 (c)
5
4 (d)
24
7
12. sin
1 cos
A
A+=?
(a) tan2
A (b) tan A (c) cot
2
A (d) cot A
13. 1
cot tan2 2 2
A A −
=?
(a) sec 2A (b) tan A (c) cosec 2A (d) cot A
14. tan cot2 2
A A+ =?
(a) sec 2A (b) tan A (c) cosec 2A (d) cot A
15. If ,2
tan t=
then 2
2
1
1
t
t
+
−is equal to
(a) cos (b) sin (c) sec (d) 2cos
16. If ,2cossin =+ AA then =A2cos
(a) 4
1 (b)
2
1 (c)
2
1 (d)
2
3
17. =− AAAA cossin2cossin2 33
(a) A4sin (b) A4sin2
1
(c) A4sin4
1 (d) None of these
18. If tan A = 1 then (sinA + cosA)2=?
(a) 1 (b) 2 (c) 1
4 (d)
1
2
19. If cos A = 1
2 then tan2A =?
(a) 3 (b) 1
3 (c) 3− (d)
1
3−
20. +
−
sin cos
sin cos
A A
A A =?
(a) 1 sin2
1 sin2
A
A
+
− (b)
1 sin2
1 sin2
A
A
−
+
Exercise - 1
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(c) sin 2A (d) 1
sin2A
21. =15cos
(a) 2
30cos1 + (b)
2
30cos1 −
(c) 2
30cos1 + (d)
2
30cos1 −
22. 2
tanA
is equal to
(a) A
A
sin1
sin1
+
− (b)
A
A
sin1
sin1
−
+
(c)A
A
cos1
cos1
+
− (d)
A
A
cos1
cos1
−
+
23. If 2cos2A = cos245°+ sin60° then find the value of cos2A
(a) 3 1
2
+ (b)
3 1
2
− (c) - 1 (d) 1
24. −
+
cot tan
cot tan
A A
A A=?
(a) sec 2A (b) sin 2A (c) cos 2A (d) cosec 2A
25.
2 2
2 2
1 1sin 22 cos 22
2 21
sin 22 cos 222 2
−
+
=?
(a) 1
2 (b) 2 (c) 1 (d)
1
2−
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1. 1 cos2 sin2
1 cos2 sin2
A A
A A
− +
+ + =?
(a) tan A (b) cot A (c) tan 2A (d) cot 2A
2. cosec 2A + cot 2A =?
(a) cot 2A (b) tan 2A (c) tan A (d) cot A
3. sin2 cos
1 cos2 1 cos
A A
A A
+ +=?
(a) tan 2A (b) cot 2A (c) tan2
A (d) cot
2
A
4. 2sin sin2
2sin sin2
+
− =?
(a) cot2
2
(b) tan2
2
(c) cot
2
(d) tan
2
5. If ,1
2
1cos
+=
aa then the value of 3cos is
(a)
+
3
3 1
8
1
aa (b)
+
aa
1
2
3
(c)
+
3
3 1
2
1
aa (d)
+
3
3 1
3
1
aa
6. Which of the following is rational
(a) 15sin (b) 15cos
(c) 15cos15sin (d) 75cos15sin
7. =++
−1
cossin
3cos3sin
(a) 2sin2 (b) 2cos2 (c) 2tan (d) 2cot
8. If ,sin
cos1tan
B
BA
−= find A2tan in terms of Btan and
show that
(a) BA tan2tan = (b) BA 2tan2tan =
(c) BBA tan2tan2tan 2 += (d) None of the above
9.
+ +
+ −
1 sin2 cos2
1 sin2 cos2=?
(a) tan (b) cot2 (c) cot (d) tan2
10. If cosA = 3
2, then cos3A =?
(a) 0 (b) 1 (c) 1
2 (d)
1
2−
11. If ,5
3sin
−= where ,
2
3 then cos
2
=
(a) 10
1− (b)
10
1 (c)
10
3 (d)
10
3−
12. If cos45° = 1
2then sin22
1
2 =?
(a) 2 2
2
+ (b)
−2 2
2
(c) 3 1
2 2
− (d) None of these
13. If ,5
1cossin =+ xx then x2tan is
(a) 17
25 (b)
25
7 (c)
7
25 (d)
7
24
14. sin2 (A + B)- sin2(A -B) =?
(a) 1 (b) sin 2A
(c) sin 2A sin 2B (d) sin A sin B
15. tan A cot2
A-1 =?
(a) sec A (b) cos A (c) cosec A (d) cos2
A
16. 3 3sin 15 cos 15
sin15 cos15
−
− =?
(a) 3
5 (b)
4
5 (c) 1 (d)
5
4
17. If cos A = 3
5 and cosB =
4
5 then find the value of cos
2
A B−
where both A and B are positive acute angles.
Exercise - 2
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(a) 25
49 (b)
24
49 (c)
49
50 (d)
49
25
18. Let .4
0
x then =− xx 2tan2sec
(a)
−
4tan
x (b)
− x
4tan
(c)
+
4tan
x (d)
+
4tan2
x
19. 1 + tan 2Atan A =?
(a) sec2A (b) sin2A (c) cot2A (d) cosec2A
20. =− A
A
sin1
cos
(a) AA tansec − (b) AA cotcosec +
(c)
−
24tan
A (d)
+
24tan
A
21. −
cos2
1 sin2
A
A=?
(a) tan (450 + A) (b) tan (450 - A)
(c) sec A – tan A (d) 1
22. sec2 A (1 + sec2A) =?
(a) cosec 2A (b) sec 2A
(c) 2 sec 2A (d) cosec A – cot A
23. 2 cos (45°+ x) cos (45°- x) =?
(a) sin x (b) cos 2x (c) sec x – cos x (d) 0
24.
tan tan4 4
tan tan4 4
+ − −
+ + −
=?
(a) tan 2α (b) sin 2α (c) cot α (d) cot 2α
25. 2 1 2 2cos8+ + + =?
(a) 2 cos (b) 2 sin (c) 2 cos2 (d) 1
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Answer key
1 a 2 c 3 b 4 b 5 c
6 d 7 a 8 c 9 b 10 b
11 d 12 a 13 d 14 c 15 a
16 b 17 b 18 b 19 c 20 a
21 a 22 c 23 b 24 c 25 d
1. (a)
sin2
1 cos2
A
A−=
( )2
2sin cos
1 1 2sin
A A
A− −
= 2
2sin cos
1 1 2sin
A A
A− +=
2
2sin cos
2sin
A A
A=
cos
sin
A
A = cot A
2. (c)
cos4A –sin4A
= (cos2A + sin2A) (cos2A – sin2A) = 1× cos 2A = cos2A
3. (b)
2
2tan15
1 tan 15
− = tan (2×15°) = tan30°=
1
3
4. (b)
(sin A - cosA)2 = sin2A + cos2A –2 sinA cosA
= 1 – sin2A
5. (c)
( )2
sin2 2sin cos
1 cos2 1 1 2sin
A A A
A A=
− − −
= 2
2sin cos
1 1 2sin
A A
A− + =
2
2sin cos
2sin
A A
A=
cos
sin
A
A = cotA
6. (d)
cotA – tanA = cos
sin
A
A -
sin
cos
A
A
= 2 2cos sin
sin cos
A A
A A
−=
cos2
sin cos
A
A A =
2cos2
2sin cos
A
A A
= 2cos2
sin2
A
A = 2cot 2A
7. (a)
sin sin2
1 cos cos2
+
+ + =
2
sin 2sin cos
1 cos 2cos 1
+
+ + −
= ( )( )
sin 1 2cos
cos 1 2cos
+
+ =
sin
cos
= tan θ
8. (c)
cosA = 5
13
sinA =
25
113
−
= 25
1169
− = 144
169=
12
13
sin 2A = 2sinAcosA = 2×12
13×
5
13=
120
169
9. (b)
tanA = 1
3
tan2A = 2
2tan
1 tan
A
A−=
2
1 21
3 311 11 3
3
= −−
=
1
32
3
= 2 3
2 3
= 3
10. (b)
cos 2A = 13
36 [ 2cos2A- 1 = cos2A]
2cos2A – 1 = 13
36 2cos2A = +
131
36=
49
36
cos2A = 49
2 36 cos A =
49
2 36=
7
6 2
11. (d)
sinA = 3
5,
p
h
Solutions
Exercise - 1
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base = 2 25 3− = 25 9− = 16 = 4
tan A = p
b=
3
4
tan 2A = 2
2tan
1 tan
A
A− =
2
32
4
31
4
−
=
3
29
116
−
=
3
27
16
= 3 16
7 2
=
24
7 =
33
7
12. (a)
sin
1 cos
A
A+
= 2
2sin cos2 2
1 2cos 12
A A
A+ −
= 2
2sin cos2 2
2cos2
A A
A=
sin2
cos2
A
A= tan
2
A
13. (d)
1
2cot tan
2 2
A A −
=
1
2
cos sin2 2
sin cos2 2
A A
A A
−
=
2 2cos sin2 2
2sin cos2 2
A A
A A
− =
cos
sin
A
A
2 2cos cos sin2 2
sin 2sin cos2 2
A AA
A AA
= −
=
= cot A
14. (c)
tan2
A+ cot
2
A =
sin2
cos2
A
A +
cos2
sin2
A
A
=
2 2sin cos2 2
cos sin2 2
A A
A A
+= 1
cos sin2 2
A A
= 2
2cos sin2 2
A A =
2
sin A= 2cosecA
15. (a)
tan2
t= ⇒
2
2
1
1
t
t
+
− =
2
2
1 tan2 cos
1 tan2
−
=
+
16. (b)
2cossin =+ AA
On squaring both the sides
oAA 90sin12sin22sin1 ===+
oA 902 = or oA 45=
Now, 2
1
2
1)45(coscos
2
22 =
== oA
17. (b)
AAAA cossin2cossin2 33 − )sin(coscossin2 22 AAAA −=
2sin cos cos2A A A= = sin2 cos2A A = 1
sin 42
A
18. (b)
tanA = 1
(sin A + cos A)2
= sin2 A + cos2A + 2sinA cosA
= 1 + sin 2A = 1 + 2
2tan
1 tan
A
A+ = 1+
2
2 1
1 1
+=2
Or
tan A = 1 A = 450
Now (sin450 + cos450)2 =
21 1
2 2
+
= ( )2
2 =2
19. (c)
cos A = 1
2, A = 600
tan A = 3
tan2A = 2
2tan
1 tan
A
A−=
( )2
2 3
1 3
−
= 2 3
1 3−=
2 3
2−= − 3
20. (a)
sin cos
sin cos
A A
A A
+
−= ( )
( )
2
2
sin cos
sin cos
A A
A A
+
−
=
2 2
2 2
sin cos 2sin cos
sin cos 2sin cos
A A A A
A A A A
+ +
+ −=
1 sin2
1 sin2
A
A
+
−
21. (a)
1 cos(2 15 )
cos152
oo + = =
1 cos30
2
o+
( )015cos o
22. (c)
sin( / 2)
tan2 cos( / 2)
A A
A
=
=
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(1 cos ) / 2 1 cos
(1 cos ) / 2 1 cos
A A
A A
− − =
+ +
23. (b)
2cos2 A = cos2 45°+sin60°
=
21
2
+ 3
2=
1
2+
3
2=
1 3
2
+
cos2A = 2cos2 A – 1
= 1 3
2
+ - 1 =
1 3 2
2
+ −=
3 1
2
−
24. (c)
−
+
cot tan
cot tan
A A
A A=
−
+
1tan
tan1
tantan
AA
AA
= −
+
2
2
1 tan
1 tan
A
A= cos 2A
25. (d)
2 2
2 2
1 1sin 22 cos 22
2 21
sin 22 cos 222 2
−
+
+ = 2 2sin cos 1
= − 2 21 1sin 22 cos 22
2 2 =
− −
2 21 1cos 22 in 22
2 2s
= - cos450 =1
2−
Answer key
1 a 2 d 3 c 4 a 5 c
6 c 7 a 8 a 9 c 10 a
11 a 12 b 13 d 14 c 15 a
16 d 17 c 18 b 19 a 20 d
21 a 22 c 23 b 24 b 25 a
1. (a)
1 cos2 sin2
1 cos2 sin2
A A
A A
− +
+ +=
( )2
2
1 1 2sin 2sin cos
1 2cos 1 2sin cos
A A A
A A A
− − +
+ − +
= 2
2
1 1 2sin 2sin cos
2cos 2sin cos
A A A
A A A
− + +
+=
2
2
2sin 2sin cos
2cos 2sin cos
A A A
A A A
+
+
= ( )( )
2sin sin cos
2cos cos sin
A A A
A A A
+
+=
sin
cos
A
A= tan A
2. (d)
cosec 2A + cot 2A
= 1 cos2 1 cos2
sin2 sin2 sin2
A A
A A A
++ =
= 21 2cos 1
2sin cos
A
A A
+ −=
22cos
2sin cos
A
A A =
cos
sin
A
A= cot A
3. (c)
sin2 cos
1 cos2 1 cos
A A
A A
+ +=
( )
++ −2
2sin cos cos
1 cos1 2cos 1
A A A
AA
= ( )2
2sin cos cos sin
1 cos2cos 1 cos
A A A A
AA A
=
+ +
= 2
2sin cos2 2
1 2cos 12
A A
A+ −
= 2
2sin cos2 2
2cos2
A A
A= tan
2
A
4. (a)
2sin sin2
2sin sin2
+
− =
2sin 2sin cos
2sin 2sin cos
+
−
= ( )( )
2sin 1 cos
2sin 1 cos
+
−
= 1 cos
1 cos
+
− =
2
2
1 2cos 12
1 1 2sin2
+ −
− −
Exercise - 1
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98
=
− +
2
2
2cos2
1 1 2sin2
=
2
2
2cos2
2sin2
= cot2
2
5. (c)
cos3 = 34cos 3cos −
= 3
3
1 1 1 14 3
22a a
a a
+ − +
=
21 1 1
32
a aa a
+ + −
= 3
3
1 1
2a
a
+
6. (c)
=−
=−=22
13)3045sin(15sin ooo irrational
22
13)3045cos(15cos
+=−= ooo =irrational
)15cos15sin2(2
115cos15sin oooo =
1 1 1 1
sin302 2 2 4
o= = = = rational
ooooo 15sin15sin15sin75cos15sin 2==
8
324
22
132
−=
−= = irrational
7. (a)
sin3 cos3
sin cos
−
+ + 1
= 3 33sin 4sin (4cos 3cos )
sin cos
− − −
+ + 1
= 3 33(sin cos ) 4(sin cos )
sin cos
+ − +
+ + 1
=2 2(sin cos ){3 4(sin sin cos cos )}
sin cos
+ − − +
++1
1)cossin1(43 +−−= 2sin2cossin4 ==
8. (a)
1 cos
tansin
BA
B
−= =
22sin ( / 2)
2sin( / 2) cos( / 2)
B
B B
= tan2
B A =
2
B, 2A = B BA tan2tan =
9. (c)
+ +
+ −
1 sin2 cos2
1 sin2 cos2 =
( )( )
+ +
− +
1 cos2 sin2
1 cos2 sin2
2
2
2cos 2sin cos
2sin 2sin cos
+=
+
( )( )
+=
+
2cos cos sin
2sin cos sin
=
=
coscot
sin
10. (a)
cos A = 3
2
sin A = =
2
31
2
−
= 3
14
− = 1
4 =
1
2
cos 3A = 4cos3A – 3cosA
= 4
3
3
2 - 3×
3
2
= 4 ×3 3 3 3 3 3 3 3
8 2 2 2− = − = 0
0r cos A = 3
2 A = 300
Now cos3A = cos 900 = 0
11. (a)
2
cos1)2/cos(
+−=
cos = 21 sin − − [ lies in IIIrd Quadrant]
5
4
25
91 −=−−=
10
1
2
5
41
)2/cos( −=
−
−=
12. (b)
cos 45° = 1
2
1- 2sin245
2
=
1
2 [ 1 – 2 sin2
2
A= cosA]
2sin2 45
2
=
11
2− 2sin2
45
2
=
2 1
2
−
2sin245
2
=
( )2 2 1
2 2
−
=
2 2
2
−
sin 221
2 =
2 2
4
− =
−2 2
2
13. (d)
5
1cossin =+ xx
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99
25
1cossin2cossin 22 =++ xxxx
25
242sin
−=x and
25
72cos
−=x
7
242tan =x
14. (c)
sin2(A + B) – sin2 (A - B)
= {sin (A + B) + sin (A - B)} {(sin (A + B) – sin(A - B)}
= (sin A cos B + cos A sin B + sin A cos B – cosA sin B) + {
sin A cos B + cos A sin B – (sin A cos B – cos A sin B)}
= (2sinAcosB) (2cosAsinB)
= (2 sin A cos A) (2 cos B sin B)
= sin 2A sin 2B
15. (a)
tan A cot2
A - 1=
2
2tan2 cot 1
21 tan
2
AA
A−
−
= −
− 2
2tan cot2 2 1
1 tan2
A A
A= −
− 2
21
1 tan2
A
=
2
2
2 1 tan2
1 tan2
A
A
− +
−
=
2
2
1 tan2
1 tan2
A
A
+
−
= 1
cos A= sec A
16. (d)
3 3sin 15 cos 15
sin15 cos15
−
−
=( )( )
( )
2 2sin15 cos15 sin 15 cos 15 sin15 cos15
sin15 cos15
− + +
−
= 1 + sin15° cos15° = 1 + 2sin15 cos15
2
= 1+ sin30
2
=
11
4+ =
5
4
17. (c)
cos A = 3
5
sin A = 21 cos A− =
23
15
−
= 4
5
and cos B = 4
5
sin B = 21 cos B− =
24
15
−
= 3
5
cos (A - B) = 2cos2−
2
A B -1
Now,
cos A cos B + sin A sin B= 2cos2−
2
A B – 1
3 4 4 3
5 5 5 5 + = 2cos2
− 2
A B – 1
2cos2 −
2
A B – 1 =
24
25
cos2 −
2
A B =
49
50 cos
− 2
A B =
7
5 2
18. (b)
sec2 tan2x x− = 1 sin 2
cos2
x
x
−
= 2
2 2
(cos sin )
(cos sin )
x x
x x
−
− =
cos sin 1 tan
cos sin 1 tan
x x x
x x x
− −=
+ +
= tan tan
4
1 tan tan4
x
x
−
+
= tan4
x −
19. (a)
1+ tan2A tanA = 1 + 2
2tan
1 tan
A
A− tanA
= 1+ 2
2
2tan
1 tan
A
A− =
2 2
2
1 tan 2tan
1 tan
A
A
− +
−
= 2
2
1 tan
1 tan
A
A
+
−=
2
2
2
2
sin1
cossin
1cos
A
AA
A
+
−
= 2 2
2 2
cos sin
cos sin
A A
A A
+
−=
1
cos2A= sec2A
20. (d)
cos
1 sin
A
A−
= 2
cos (1 sin )
cos
A A
A
+ =
(1 sin )
cos
A
A
+
2
cos sin2 2
cos sin cos sin2 2 2 2
A A
A A A A
+
=
+ −
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100
= cos sin
2 2
cos sin2 2
A A
A A
+
−2
tan1
2tan1
A
A
−
+
=
+=
24tan
A
21. (a)
cos2
1 sin2
A
A− =
2 2
2 2
cos sin
cos sin 2sin cos
A A
A A A A
−
+ −
= ( )( )
( )2
cos sin cos sin
cos sin
A A A A
A A
+ −
−
= cos sin
cos sin
A A
A A
+
−=
+
−
1 tan
1 tan
A
A= tan (450 + A)
22. (c)
sec2 A (1+sec2A) = 2
1
cos A
+
11
cos2A
= 2
1
cos A
+
cos2 1
cos2
A
A=
2
1
cos A
− +
22cos 1 1
cos2
A
A
= 2
2
1 2cos
cos2cos
A
AA=
2
cos2A = 2sec2A
23. (b)
2cos (45° + x) cos (45° - x)
= 2(cos45°cosx – sin450sinx) (cos45°cosx + sin45°sinx)
= 2 1 1
cos sin2 2
x x
−
1 1cos sin
2 2x x
+
= 2×1 1
2 2 (cos x – sin x) (cos x + sin x)
= 2
2 (cos2x – sin2x) = cos2x
24. (b)
tan tan4 4
tan tan4 4
+ − −
+ + −
=
+ −−
− ++ −
+− +
1 tan 1 tan
1 tan 1 tan1 tan 1 tan
1 tan 1 tan
=
( ) ( )( )( )
( ) ( )( )( )
+ − −
+ −
+ + −
+ −
2 2
2 2
1 tan 1 tan
1 tan 1 tan
1 tan 1 tan
1 tan 1 tan
= ( ) ( )
( ) ( )
+ − −
+ + −
2 2
2 2
1 tan 1 tan
1 tan 1 tan
=
+ 2
2tan
1 tan= sin 2α
25. (a)
2 2 2(1 cos8 )+ + +
= 22 2 2(2cos 4 )+ + 2 8
1 cos8 2cos2
+ =
= ( )22 2 4cos 4+ + = ( )2 2 1 cos4+ + +
= ( )22 2 2cos 2+ = 2 2cos2+ = ( )2 1 cos2+
= ( )22 2cos = 2 cos
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Some Important Results
1. cos A cos (60° – A) cos (60° + A = 1
4cos 3A
2. sin A sin (60° – A) sin (60° + A) = 1
4sin 3A
3. tan tan (60° – ) tan (60° + ) = tan 3
4. sin2 + sin2(600 - ) + sin2(600 + ) = 3
2
5. cos2 + cos2(600 - ) + cos2(600 + ) = 3
2
6. cos cos 2 cos 4 ……… cos2n-1 = n
n
sin2
2 sin
7. sin + sin (+) + sin (+) +.... + sin (+n−) =
sin2 sin ( 1)
2sin
2
+ −
n
n
1. cos A cos (60° – A) cos (60° + A = 1
4cos 3A
Proof:
cos A cos (60° – A) cos (60° + A)
= cos A (cos2 60° – sin2 A)
[ cos (A + B) cos (A – B) = cos2 A – sin2 B]
= ( )2 21 1cosA sin A cosA 1 cos A
4 4
− = − −
= 23cosA cos A
4
− +
= 1
4cos A (–3 + 4 cos2 A)
= 1
4 (4 cos3 A – 3 cos A)
= 1
4cos 3A
2. sin A sin (60° – A) sin (60° + A) = 1
4sin A
Proof:
sin A sin (60° – A) sin (60° + A)
= sin A (sin2 60° – sin2 A)
[ sin (A + B) sin (A – B) = sin2 A – sin2 B]
= ( )2 23 1sinA sin A sinA 3 4sin A
4 4
− = −
= 1
4 (3 sin A – 4 sin3 A)
= 1
4sin 3A
3. tan tan (60° – ) tan (60° + ) = tan 3
4. sin2 + sin2(600 - ) + sin2(600 + ) = 3
2
Proof:
sin2 + sin2 (60° – ) + sin2 (60° + )
= sin2 + (sin 60° cos – cos 60° sin )2 +
(sin 60° cos + cos 60° sin )2
= sin2 + 2 2
3 1 3 1cos sin cos sin
2 2 2 2
− + +
= sin2 + 2 2 2 23 1 3 1
cos sin cos sin4 4 4 4
+ + +
= 2 23 3 3
sin cos2 2 2
+ =
5 cos2 + cos2 (60° – ) + sin2 (60° + ) =3
2
Proof:
cos2 + cos2 (60° – ) + sin2 (60° + )
= cos2 + (cos 60° cos + sin 60° sin )2 +
(cos 60° cos – sin 60° sin )2
= cos2 +
2 2
cos 3 cos 3sin sin
2 2 2 2
+ + −
= cos2 + 2 2
2cos 3 3 cossin sin cos
4 4 2 4
+ + +
23 3sin sin cos
4 2+ −
= 2 23 3 3
cos sin2 2 2
+ =
6. cos cos 2 cos 4 ……… cos2n-1 = n
n
sin2
2 sin
Proof:
Special Properties 9
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102
cos cos 2 cos 4 ……… cos2n-1
= 1
22sin
sin cos cos 2 cos 4 …… cos2n-1
= 2
1
2 sin2 sin 2 cos 2 cos 4 ……… cos2n-1
= 3
1
2 sin2 sin 4 cos 4 ……… cos2n-1
Similarly
= n 1
1
2 sin− sin2n-1 cos2n-1
= n
1
2 sin2sin2n-1 cos2n-1
= n
1
2 sinsin (2×2n-1) =
n
n
sin2
2 sin
Some Standard Identities in Triangle
1. tan A + tan B + tan C = tan A tan B tan C
2. tan2
Atan
2
B+ tan
2
Ctan
2
B+ tan
2
Ctan 1
2
A=
3. sin 2A +sin 2B + sin 2C = 4 sin A sin B sin C
4. cos 2A + cos 2B + cos 2C = -1 – 4cos A cos Bcos C
5. cos A + cos B + cos C = 1 + 4 sin 2
Asin
2
Bsin
2
C
6. sin A + sin B + sin C = 4 cos 2
A cos
2
Bcos
2
C
1. tan A + tan B + tan C = tan A tan B tan C
Proof: in Δ ABC, we have A + B + C =
A + B = - C tan (A+B) = tan (-c)
tan tan
1 tan tan
A B
A B
+=
−- tan C
tan A + tan B = - tan C + tan A tan B tan C
tan A + tan B + tan C = tan A tan B tan C
2. tan2
Atan
2
B+ tan
2
Ctan
2
B+ tan
2
Ctan 1
2
A=
Proof:
Since A + B + C = , we have 2 2 2 2
A B C+ = −
tan tan cot
2 2 2 2 2
A B C C + = − =
tan tan
12 2
1 tan tan tan2 2 2
A B
A B c
+
=
−
tan2
Atan
2
C+ tan
2
Btan
2
C = 1 - tan
2
Atan
2
B
3. sin 2A +sin 2B + sin 2C = 4 sin A sin B sin C
Proof : (sin2A+sin 2B) + sin 2C = 2 sin (A+B) cos (A-B) + sin 2C
= 2 sin (-C) cos (A-B) + sin 2C
= 2 sin C cos (A-B) + 2 sin C cos C
= 2 sin C [cos (A-B) + cosC]
= 2 sin C [cos (A-B) + cos {-(A+B)]
= 2sin C[cos (A-B) – cos (A+B)]
= 2 sin C ×2 sin A sin B = 4 sin A sin B sinC
4. cos 2A + cos 2B + cos 2C = -1 – 4cos A cos Bcos C
Proof: (cos 2A + cos 2B) + cos 2C
= 2 cos (A+B) cos (A-B) + 2cos2 C-1
=2 cos (-C) cos (A-B) + 2cos2 C – 1
= -2 cos C cos (A-B) + 2cos2 C-1
=- 2 cos C [cos (A-B) –cos C]-1
=- 2 cos C [cos (A-B) –cos {-(A+B)} ]-1
= - 2 cos C [cos (A-B) + cos (A+B)] – 1
= - 1-4 cos A cos B cos C
5. cos A + cos B + cos C = 1 + 4 sin 2
Asin
2
Bsin
2
C
Proof:
(Cos A + cos B) + cos C -1
= 2cos cos cos2 2
A B A B+ −+ C – 1
= 2cos 2 2
C −
cos 2
A B−+ cos C – 1
= 2sin 2
C cos
2
A B−+1 – 2sin2
2
C - 1
= 2sin 2
Ccos
22sin2 2
A B C−−
= 2sin 2
Ccos sin
2 2
A B C− −
= 2sin 2
Ccos sin
2 2 2
A B A B − + −
= 2sin 2
Ccos cos
2 2
A B A B− + −
= 2sin 2
C2sin sin 4
2 2
A B =
sin
2
Acos
2
Bsin
2
C
6. sin A + sin B + sin C = 4 cos 2
A cos
2
Bcos
2
C
Proof :
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103
(sin A + sin B) + sin C
= 2sin cos2 2
A B A B+ −+ sin C
= 2sin cos sin2 2 2
C A B − − +
C
= 2cos 2
Ccos
2
A B− + 2sin cos
2 2
C C
= 2cos 2
Ccos sin
2 2
A B C− +
= 2cos cos sin2 2 2 2
C A B A B − + + −
= 2cos cos cos2 2 2
C A B A B− + +
= 4cos cos cos2 2 2
A B C
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104
1. cos 50° cos 70° cos 10° =?
Sol. cos 10°cos 50° cos 70°
= cos 100 cos( 600 – 100) cos(600 + 100)
= 1
4 cos 30°=
1 3 3
4 2 8 =
2. sin5 sin65 sin55
cos75
=?
Sol. sin5 sin65 sin55
cos75
=
( ) ( )sin5 sin 60 5 sin 60 5
cos75
+ −
( )sin 3 51
4 cos75
=
=
1 sin15
4 cos75
=
1 sin15 1
4 sin15 4
=
3. tan 12° cot 18° tan 48° cot 54°=?
Sol. tan 12° tan 48° cot 18° cot 54°
= tan12° tan (60°-12°) tan (60°+12°) cot 54°
= tan (3 × 120) cot 54°
= tan 36° cot 54° = 1
4. sin2 10° + sin2 70° + sin2 50° =?
Sol. sin2 10° + sin2 70° + sin2 50°
= sin2 10° + sin2 (60° + 10°) + sin2 (60° 10°) = 3
2
5. cos2 23° + cos2 83° + cos2 37° + sin2 60°
Sol. cos2 23° + cos2 83° + cos2 37° + sin2 60°
= cos223°+ cos2 (60°+23°) + cos2 (60° - 23°) + sin2 60°
=
2
3 3
2 2
+
= 3 3 9
2 4 4+ =
6. cos 20° cos 40° cos 60° cos 80°
Sol. (cos 20° cos 40° cos 80°) cos 60°
1st method:
= 3 4
sin160 1 1 1
2 162 sin20 2
= =
2nd method:
= cos3 20 1
4 2
=
cos60 1 1
4 2 16
=
7. 2 3
cos cos cos7 7 7
Sol. 2 3cos cos cos
7 7 7
= 2
42sin
37 cos7
2 2 sin7
=
4 3 4 3sin sin
7 7 7 7
8sin7
+ + −
=
sin sin178
8sin7
+
=
8. Find the value of cos 2 4 6
cos cos7 7 7
+ +
Sol. S = cos 2 4 6
cos cos7 7 7
+ +
=
sin 337
cos7 7
sin7
+
=
3 4 72sin cos sin sin
17 7 7 7
2 2 22sin 2sin
7 7
−
= = −
9. sin + sin 3 + sin 5 +..... + sin(2n-1) =?
Sol. sin + sin3 + sin5 +....+sin(2n-1)
=
2sin
2 (2 1)sin
2 2sin
2
nn
+ −
= 2sin
sin
n
10. If A + B + C = 180°,then cos2A +cos2B +cos2C=
Sol. 2 2 2cos cos cosA B C+ +
= 1 cos2 1 cos2 1 cos2
2 2 2
A B C+ + ++ +
= ( )1 3
cos2 cos2 cos22 2
A B C= + + +
( )1 3
1 4cos cos cos2 2
A B C= − − +
= 1 – 2cosA cos B cos C
11. In triangle ABC, cos2A+ cos2B-cos2C =?
Sol. 2 2 2cos cos cosA B C+ −
Examples
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105
= 2 2 2cos sin sinA C B+ −
= ( ) ( )2cos sin inA C B s C B+ + −
= 1-sin2A + sin A sin (C -B)
= 1-sin A [sinA – sin(C-B)]
= 1-sin A [sin(B+C)-sin(C-B)] = 1-2 sin A sin B cosC
12. In triangle ABC,
sin (B+C-A) +sin (C + A - B) + sin (A + B -C) =?
Sol. sin (B + C – A ) + sin (C + A - B) + sin (A + B - C)
= sin ( 2A − )+ sin( 2B − ) + sin ( 2C − )
= sin 2A + sin 2B + sin 2C
= 4 sin A sin B sin C
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106
1. tan 20° tan 80° cot 50° = ?
(a) 3 (b) 1
3 (c) 2 3 (d)
1
2 3
2. Find the value of 1 – sin 10° sin 50° sin 70°
(a) 5
8 (b)
7
8 (c)
1
4 (d)
3
4
3. Find the value of cos 20° cos 40° cos 60° cos 80°
(a) 1
16 (b)
1
8 (c)
3
16 (d)
1
4
4. 1 1
cos15 cos7 cos822 2
=?
(a) 1
8 (b)
1
4 (c)
3
8 (d)
1
2
5. tan 6° tan 42° tan 66° tan 78° = ?
(a) 1 (b) 2 (c) 3 (d) 4
6. Find the value of
sin 20° sin 40° sin 60° sin 80°
(a) 1
16 (b)
2
16 (c)
3
16 (d)
1
4
7. sin 12° sin 48° sin 54° = ?
(a) 1
4 (b)
3
4 (c)
3
8 (d)
1
8
8. cot 15° cot 25° cot 35° cot 85° = ?
(a) 1 (b) 0 (c) 2 (d) –1
9. Find the value of cos 10° cos 30° cos 50° cos 70°
(a) 1
16 (b)
2
16 (c)
3
16 (d)
1
4
10. Find the value of sin 6° sin 42° sin66° sin 78°
(a) 1
16 (b)
1
8 (c)
3
16 (d)
1
4
11. If ,180ozyx =++ then zyx 2cos2cos2cos −+ is equal to
(a) 4sin sin sinx y z (b) 1 4sin sin cosx y z−
(c) 4 sin x sin y sin z - 1 (d) cos A cos B cos C
12. If ,2 =++ then
(a) 2
tan2
tan2
tan2
tan2
tan2
tan
=++
(b) 12
tan2
tan2
tan2
tan2
tan2
tan =++
(c) 2
tan2
tan2
tan2
tan2
tan2
tan
−=++
(d) None of these
13. A, B, C are the angles of a triangle, then
=−++ CBACBA coscoscos2sinsinsin 222
(a) 1 (b) 2 (c) 3 (d) 4
14. If A, B, C are angles of a triangle, then
CBA 2sin2sin2sin −+ is equal to
(a) CBA coscossin4 (b) Acos4
(c) AA cossin4 (d) CBA sincoscos4
15. If ,180oCBA =++ then the value of
2cot
2cot
2cot
CBA++ will be
(a) 2
cot2
cot2
cot2CBA
(b) 2
cot2
cot2
cot4CBA
(c) 2
cot2
cot2
cotCBA
(d) 2
cot2
cot2
cot8CBA
16. sin2 18 + sin2 78 + sin2 42 = ?
(a) 1
2 (b)
3
2 (c)
2
3 (d)
5
2
17. cos2 + cos2 (60° – ) – sin2 (60° + ) = ?
(a) 2
3 (b) 1 (c)
3
2 (d)
1
2
18. 2 + sin2 (60° – ) + cos2 (30° – ) – 2
1
sec =?
(a) 3
2 (b)
5
2 (c) 1 (d)
3
2−
19. sin 12° sin 24° sin 48° sin 84°
Exercise - 1
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107
(a) 1
8 (b)
7
8 (c)
1
16 (d)
9
16
20. sin2 7 + sin2 67 + sin2 53 = ?
(a) 1 (b) 1
2 (c)
5
2 (d)
3
2
21. 2
1
sin 36 – cot2 12° cot2 48° cot2 72° = ?
(a) 3 (b) 1 (c) –1 (d) 2
22. cos20 cos40 cos80 =?
(a) 1/2 (b) 1/4 (c) 1/6 (d) 1/8
23. sin2 29° + sin2 31° + sin2 89° + sin2 45° =?
(a) 3/2 (b) 1/4 (c) 2 (d) 1/8
24. If =++ CBA and ,coscoscos CBA = then CB tantan
is equal to
(a) 2
1 (b) 2 (c) 1 (d)
2
1−
25. If A, B, C, D are the angles of a cyclic quadrilateral, then
cos A + cos B + cos C + cos D =?
(a) 4 (b) 1 (c) 0 (d) –1
26. If 2 2 2cos cos cos 1A B C+ + = then, ABC is
(a) Equilateral (b) isosceles
(c) Right angled (d) none of these
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1. 15
16cos
15
8cos
15
4cos
15
2cos
=
(a) 1/2 (b) 1/4 (c) 1/8 (d) 1/16
2. )120(cos)120(coscos 222 −+++ =?
(a) 3/2 (b) 1 (c) 1/2 (d) 0
3. The value of 12
5cos
4cos
12cos 222
++ is
(a) 2
3 (b) 3
2 (c) 2
33 + (d)
33
2
+
4. The value of 16
7sin
16
5sin
16
3sin
16sin
is
(a) 16
1 (b)
16
2 (c)
8
1 (d)
8
2
5. ,, are real numbers satisfying =++ . The
minimum value of sinsinsin ++ is
(a) Zero (b) – 3 (c) Positive (d) Negative
6. If A, B, C, D are the angles of a cyclic quadrilateral, then
=+++ DCBA coscoscoscos
(a) 2(cos cos )B C+ (b) )cos(cos2 BA +
(c) )cos(cos2 DA + (d) 0
7. If ABCD is a cyclic quadrilateral, then the value of
=−+− DCBA coscoscoscos ?
(a) 0 (b) 1
(c) )cos(cos2 DB − (d) )cos(cos2 CA −
8. If CBA coscoscos = and ,=++ CBA then the value of
CB cotcot is
(a) 1 (b) 2 (c) 3
1 (d)
2
1
9. If ,180oCBA =++ then the value of )cot(cot CB +
)cot(cot)cot(cot BAAC ++ will be
(a) CBA secsecsec (b) CBA coseccoseccosec
(c) CBA tantantan (d) 1
10. If ,180oCBA =++ then =2
tan2
tanBA
(a) 0 (b) 1 (c) 2 (d) 3
11. If )0,,( =++ CBACBA and the angle C is obtuse then
(a) 1tantan BA (b) 1tantan BA
(c) 1tantan =BA (d) None of these
12. If A, B, C are acute positive angles such that
=++ CBA and ,cotcotcot KCBA = then
(a) 33
1K (b)
33
1K (c)
9
1K (d)
3
1K
13. If ,2
3=++ CBA then =++ CBA 2cos2cos2cos
(a) CBA coscoscos41 − (b) CBA sinsinsin4
(c) CBA coscoscos21 + (d) CBA sinsinsin41 −
14. In a ABC, tan A + tan B + tan C = 6 and tan A tan B = 2
then find the value of tan C.
(a) 1 (b) 2 (c) 3 (d) –3
15. In a ABC if cos2 A + cos2 B + cos2 C = 1, then ABC is
(a) equilateral (b) isosceles
(c) right angled (d) N.O.T.
16. sin 6° sin 42° sin 66° sin 78° =?
(a) 1
8 (b)
1
16 (c)
1
16 2 (d)
1
8 2
17. If A + B + C = , then sin2 2sin
2
A+ 2sin
2
B−
2
C=?
(a) 1 – 2 cos cos sin2 2 2
A B C
(b) 1 – 2 cos cos sinA B C
(c) 1 – 2 cos2 2 2cos sin2 2 2
A B C
(d) None of these
Exercise - 2
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18. 3 5 7
sin sin sin sin16 16 16 16
(a) 1
8 (b)
1
16 (c)
1
16 2 (d)
1
8 2
19. 2 3 4
sin sin sin sin5 5 5 5
=?
(a) 1
8 (b)
5
16 (c)
1
16 (d)
3
8−
20. If ,40cos20cos10cos =x then the value of x is
(a) 10tan4
1 (b) 10cot
8
1
(c) 10cosec8
1 (d) 10sec
8
1
21. 2 4
cos cos cos7 7 7
=?
(a) 0 (b) 2
1 (c)
4
1 (d)
8
1−
22. 2 4 8
cos cos cos cos5 5 5 5
=?
(a) 1/16 (b) 0 (c) – 1/8 (d) –1/16
23. If A, B and C are the angles of a plain triangle and
3
2
2tan,
3
1
2tan ==
BA then 2
tanC
=?
(a) 7/9 (b) 2/9 (c) 1/3 (d) 2/3
24. The value of tan 6° tan42° tan 66° tan 78° is
(a)1 (b) 1
2 (c)
1
4 (d)
1
8
25. Given that a, b, c are the sides of a triangle ABC which is
right angled at C, the then minimum value of 2
c c
a b
+
is
(a) 0 (b) 4 (c) 6 (d) 8
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Answer key
1 a 2 b 3 a 4 a 5 a
6 c 7 d 8 a 9 c 10 a
11 b 12 a 13 b 14 d 15 c
16 b 17 d 18 b 19 c 20 d
21 b 22 d 23 c 24 b 25 c
26 c
1. (a)
tan 20° tan 80° cot 50°
= tan 20° tan 80° tan 40°
= tan (3 × 20) = tan 60° = 3
2. (b)
1 – sin 10° sin 50° sin 70°
= 1 – 1
4. sin (3 × 10) =
1 1 71 .
4 2 8− =
3. (a)
cos 20° cos 40° cos 60° cos 80°
= 1 1
2 4 cos (3 × 20) =
1 1 1 1
2 4 2 16 =
4. (a)
1 1
cos15 cos7 cos822 2
= 1 1
cos15 cos7 sin72 2
= 1 1 1cos15 2cos7 sin7
2 2 2
= 2cos15 sin15
2 2
= ( )1 1 1 1
sin 2 154 4 2 8
= =
5. (a)
tan6 tan66 tan54 tan42 tan78
tan54
= ( )tan 3 6 tan42 tan78
tan54
=
( )tan 3 18 tan541
tan54 tan54
= =
6. (c)
sin 20° sin 40° sin 60° sin 80°
sin 60° sin 20° sin 40° sin 80°
Using sin sin (60° – ) sin (60° + ) = 1
4sin 3
sin 60° 1
4 sin (3 × 20) =
3 1 3 3
2 4 2 16 =
7. (d)
sin12 sin48 sin72 sin54
sin72
=
( )sin 3 12 sin541
4 sin72
= sin36 sin54
4sin72
=
1 sin36 cos36
4 2sin36 cos36
=
1
8
8. (a)
cot 15° cot 25° cot 35° cot 85° = cot 15° cot (3 × 250)
= cot 15° cot 75° = cot 15° tan 15° = 1
9. (c)
cos 10° cos 30° cos 50° cos 70°
cos 30° cos 10° cos 50° cos 70°
Using cos cos (60° – ) cos (60° + ) = 1
4 cos 3
= cos 30° × 1
4 cos (3 × 10) =
3 1 3 3
2 4 2 16 =
10. (a)
sin 6° cos 48° cos 24° cos 12°
sin 6° cos 12° cos 24° cos 48°
Using cos cos 2 cos 4 …… cos2n-1
= n
n
sin2
2 sin
= sin 6°.
( )3
3
sin2 12
2 sin12 =
sin6 .sin96
8.sin12
= 2 sin6 cos6 sin12 1
16 sin12 16 sin12 16
= =
11. (b)
( )cos2 cos2 cos2x y z+ −
= 22cos( )cos( ) (2cos 1)x y x y z+ − − −
1cos2)cos()cos(2 2 +−−+= zyxyx
1cos2)cos()cos(2 2 +−−−= zyxz
= 22cos cos( ) 2cos 1z x y z− − − +
=1 2cos {cos( ) cos }z x y z− − +
Solutions
Exercise - 1
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)}cos(){cos(cos21 yxyxz +−−−=
( )1 2cos 2sin sinz x y= − = 1 4sin sin cosx y z−
12. (a)
2 + + = , 2 2 2
+ + =
tan2 2 2
+ +
= tan = 0
tan2 2 2
+ + =
tan tan tan tan tan tan2 2 2 2 2 2
1 tan tan tan tan tan tan2 2
2 2 2 2
+ + −
− − −
tan tan tan tan tan tan
2 2 2 2 2 2 0
1 tan tan tan tan tan tan2 2 2 2 2 2
+ + −
=
− − −
tan tan tan2 2 2
+ + - tan tan tan
2 2 2
= 0
2
tan2
tan2
tan2
tan2
tan2
tan
=++
13. (b)
CBA 222 sinsinsin ++
CBA 222 sincos1cos1 +−+−=
)cos()cos(cos2 2 CBCBA −+−−=
)]cos([coscos2 CBAA −−−=
)]cos()cos([cos2 CBCBA −−+−−=
( )2 cos 2cos cosA B C= +
2coscoscos2sinsinsin 222 =−++ CBACBA
14. (d)
CBA 2sin2sin2sin −+
= )sin()cos(2cossin2 CBCBAA −++
),cos()cos(,,{ ACBACBCBA −=+−=+=++
}sin)sin(,cos)cos( ACBACB =+−=+
= 2sin cos 2cos sin( )A A A B C− −
)]sin([sincos2 CBAA −−=
)]sin()[sin(cos2 CBCBA −−+=
2cos 2cos sinA B C= 4cos cos sinA B C=
15. (c)
oCBA 180=++
2
9022
CBA o −=+
−=
+
290cot
22cot
CBA o
cot cot 1
12 2 tan2
cot cot cot2 2 2
A B
C
B A C
−
= =
+
2
cot2
cot2
cot12
cot2
cotABCBA
+=
−
cot cot cot cot cot2 2 2 2 2
A B C C B= + cot
2
A+
16. (b)
Value of cos2 + cos2 (60° – ) + cos2 (60° + ) is always
3
2 irrespective of the value of .
So required answer is 3
2
17. (d)
cos2 + cos2 (60° – ) – (1 – sin2 (60° + ))
cos2 + cos2 (60° – ) + cos2 (60° + ) – 1 = 3 1
12 2
− =
18. (b)
1 + 1 – 2
1
sec + sin2 (60° – ) + cos2 (30° – )
= 1 + 1 – cos2 + sin2(60°– ) + cos2 [900– (60 + )]
= 1 + sin2 + sin2 (60° – ) + sin2 (60° + )
= 1 + 3 5
2 2=
19. (c)
sin12 sin48 sin72 sin24 sin84
sin72
= ( )sin 3 12 sin24 sin841
4 sin72
= 1 sin24 sin36 sin84
4 sin72
=
( )sin 3 241 1
4 4 sin72 16
=
20. (d)
Value of sin2 + sin2 (60° – ) + sin2 (60° + ) is always
3
2 irrespective of the value of . So required answer is
3
2.
21. (b)
cosec2 36° – (cot 12°. cot 48°. cot 72°)2
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= cosec2 36° – cot2 (3 × 12) = cosec2 36° – cot2 36 = 1
22. (d)
cos20 cos40 cos80o o o
= cos20O cos (60O – 20O) cos( 60O + 200)
= 1
cos(3 20 )4
= 1
cos604
= 1
8
23. (c)
sin2 29° + sin2 31° + sin2 89° + sin2 45°
= sin2 29° + sin2 (60° – 29°) + sin2 (60° + 29°) + sin2 45°
= 3
2 + sin2 45° =
3 1 4
2 2 2+ = = 2
24. (b)
cos cos cosA B C=
CBCB coscos)](cos[ =+−
CBCB coscos)cos( =+−
CBCBCB coscos]sinsincos[cos =−−
CBCB coscos2sinsin = 2tantan =CB
25. (c)
ABCD is a cyclic quadrilateral
A + C = 180°, B + D = 180°
C = 180° – A, D = 180° – B
cos C = cos (180° – A) = –cos A, cos D = cos (180° – B)
= – cos B
cos A + cos C = 0, cos B + cos D = 0
cos A + cos B + cos C + cos D = 0
26. (c)
2 2 2cos cos cos 1A B C+ + =
cos2A + cos2B – (1 – cos2C) = 0
cos2A + cos2B – sin2C = 0
cos2A + cos (B + C) cos (B – C) = 0
2 cos A cos B cos C = 0
Hence, either A or B or C is 90°
Answer key
1 d 2 a 3 a 4 b 5 c
6 d 7 a 8 d 9 b 10 b
11 b 12 a 13 d 14 c 15 c
16 b 17 a 18 d 19 b 20 b
21 d 22 d 23 a 24 a 25 d
1. (d)
15
16cos
15
8cos
15
4cos
15
2cos
4
4
2sin 2
152
2 sin15
= =
32sin
152
16sin15
=
sin 215
216sin
15
+
=
2sin
1 115216 16
sin15
=
2. (a)
)120(cos)120(coscos 222 oo −+++
22 )120(cos)120(coscos oo −+++=
)120(cos)120(cos2 oo −+−
= 22 2 2cos 2cos cos120 2 cos sin 120o o + − −
o120sin2cos2coscos 2222 +−+=
2
3
4
32120sin2 2 === o
3. (a)
12
5cos
4cos
12cos 222
++
+
+
−=
12
5cos
2
1
12sin1 2
2
2
−++=
12sin
12
5cos
2
11 22
3 5 5
cos cos2 12 12 12 12
= + + −
3
cos cos2 2 3
= +
3 1 30
2 2 2= + =
4. (b)
3 5 7
sin sin sin sin16 16 16 16
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1 3 5 7
2sin sin 2sin sin4 16 16 16 16
=
−
−=
4
3cos
8cos
4cos
8cos
4
1
+
−=
2
1
8cos
2
1
8cos
4
1
−=
−= 1
8cos2
8
1
2
1
8cos
4
1 22
16
2
2
1
8
1
4cos
8
1==
=
5. (c)
When =++
)sin(sinsinsin ++++
0sinsinsin ++
The minimum value of sinsinsin ++ is always
positive.
6. (d)
ABCD is a cyclic quadrilateral
So CACA −==+ 180180
CCA cos)180cos(cos −=−=
0coscos =+ CA .....(i)
Similarly, 0coscos =+ DB ..... (ii)
Adding, + + + =cos cos cos cos 0A B C D
7. (a)
ABCD is a cyclic quadrilateral
+ = 180A C CA −= 180
CCA cos)180cos(cos −=−=
0coscos =+ CA .....(i)
Now ,180=+ DB then 0coscos =+ DB ....(ii)
Subtracting (ii) from (i), we get
0coscoscoscos =−+− DCBA
8. (d)
CBA coscoscos =
ACBCBA −=+=++
cos( ) cos( )B C A + = − cos( ) cosB C A+ = −
cos cos sin sin cos cosB C B C B C− = −
( Given cos cos cos )A B C=
CBCB sinsincoscos2 =
cos cos 1
sin sin 2
B C
B C =
1cot cot
2B C =
9. (b)
cot cotB C+
= sin cos sin cos
sin sin
C B B C
B C
+
CB
CB
sinsin
)sin( +=
CB
Ao
sinsin
)180sin( −=
CB
A
sinsin
sin=
Similarly, AC
BAC
sinsin
sincotcot =+
and BA
CBA
sinsin
sincotcot =+
Now,
)cot)(cotcot)(cotcot(cot BAACCB +++
sin sin sin
sin sin sin sin sin sin
A B C
B C C A A B=
=cos cos cosecA ecB ecC
10. (b)
oCBA 180=++
+−=
222
CBA
+=
22tan
2cot
CBA
2tan
2tan1
2tan
2tan
2tan
1
CB
CB
A−
+
=
1 tan tan tan tan tan tan2 2 2 2 2 2
B C A B A C− = +
tan tan tan tan tan tan 12 2 2 2 2 2
A B B C A C+ + =
i.e. , = 12
tan2
tanBA
11. (b)
CBACBA −=+=++ )tan()tan( CBA −=+
)tan(tantan1
tantanC
CA
BA−=
−
+ C
BA
BAtan
tantan1
tantan−=
−
+
Now C is an obtuse angle,
hence
0tan0tan − CC
0tantan10tantan1
tantan−
−
+ BA
BA
BA
BA,( are acute angles; 0tan,0tan BA )
1tantan BA
12. (a)
=++ CBA
CBACBA tantantantantantan =++
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Now A.M. G.M.
3/1)tantan(tan
3
tantantanCBA
CBA
++
3/1)tantan(tan3
tantantanCBA
CBA
3)tantan(tan 3/2 CBA
33
13
13
1 2/3
3/2
K
KK
13. (d)
CBA 2cos2cos2cos ++
CBABA 2cos)cos()cos(2 +−+=
CBAC 2cos)cos(2
3cos2 +−
−=
CBAC 2sin21)cos(sin2 −+−−=
}sin){cos(sin21 CBAC +−−=
+−+−−= )(
2
3sin)cos(sin21 BABAC
)}cos(){cos(sin21 BABAC +−−−=
CBA sinsinsin41 −= .
14. (c)
In a ABC we know that
A + B + C = and
tan A + tan B + tan C = tan A tan B tan C
6 = 2 tan C
tan C = 3
15. (c)
cos2 A + cos2 B + 1 – sin2 C = 1
cos2 A + cos2 B – sin2 C = 0
cos2 A + cos (B + C) cos (B – c) = 0
cos2 A – cos A cos (B – C) = 0
( A + B + C = B + C = – A)
cos2 A = cos A cos (B – C)
cos A = cos (B – C) A = B – C A + C = B
We know that A + B + C = 2B =
B = 2
Hence is right angled.
16. (b)
Ist method
cos 84° cos 48° cos 24° cos 12°
= 2sin96
cos842 8sin12
=( ) ( )sin 96 84 sin 96 84
16sin12
+ + −
= sin180 sin12 1
16sin12 16
+ =
IInd method:
= sin6 sin66 sin54 sin18 sin42 sin78
sin54 sin18
=
sin18 sin5414 4
sin54 sin18 16
=
17. (a)
sin2 2sin
2
A+ 2sin
2
B−
2
C
= sin sin2 2
A C A C+ −
+ 1-2cos
2
B
= cos 2sin cos 12 2 2
B A C B− − +
= cos sin cos 12 2 2
B A C B − − +
= cos sin sin 12 2 2
B A C A C − + − +
= 1 – 2 cos cos sin2 2 2
A B C
18. (d)
3 5 7sin sin sin sin
16 16 16 16
= 3 3sin sin sin sin
16 16 2 16 2 16
− −
= 3 3
sin sin cos cos16 16 16 16
=
1 3sin sin
4 8 8
= 1sin sin
4 8 2 8
−
=1
sin cos4 8 8
=
1 1sin
4 2 4
=
1
8 2
19. (b)
2 3 4
sin sin sin sin5 5 5 5
= 2 2sin sin sin sin
5 5 5 5
− −
=
22
sin sin5 5
=
21 2
2sin sin2 5 5
=
2
1 2 2cos cos
4 5 5 5 5
− − +
=
21 3
cos cos4 5 5
−
[∵ cos(-) = cos ]
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= ( )21
cos36 cos1084
− =
2
1 5 1 5 1
4 4 4
+ −+
=
2
1 5 5
4 2 16
=
20. (b)
ooox 40cos20cos10cos=
1
[2 sin10 cos10 cos 20 cos 40 ]2 sin10
o o o o
o=
]40cos20cos20sin2[10sin2.2
1 ooo
o=
1
[2sin40 cos40 )2.4sin10
o o
o= =
1(sin80 )
8sin10o
o
oo
o10cot
8
110cos
10sin8
1==
21. (d)
2 4
cos cos cos7 7 7
3
3
sin 2 .7
2 sin7
=
=
8sin
7
8sin7
=
sin7
8sin7
+
=
sin17
88sin
7
−
= −
22. (d)
5
8cos
5
4cos
5
2cos
5cos
=
4
4
2sin
5
2 sin5
=
16sin
5
16sin5
=
sin 35
16sin5
+
= sin
5
16sin5
−
= 1
16−
23. (a)
=++ CBA
−=
+
22tan
2tan
CBA
tan tan
2 2 cot2
1 tan .tan2 2
A BC
A B
+
=
−
1 293 3 cot
1 2 7 21
3 3
C+
= =
−
9
7
2tan =
C
24. (a)
tan 6° tan 42° tan 66° tan 78°
= tan6° tan(60° – 18°) tan(60° + 6°) tan(60° + 18°)
= ( ) ( ) ( )tan6 tan 60 6 tan18 tan 60 18 tan 60 18
tan18
+ − +
= ( ) ( )tan6 tan 60 6 tan 3 18
tan18
+
= ( ) ( )tan6 tan 60 6 tan 60 6 tan18
1tan18 tan18
− + = =
25. (d)
A b C
a c
B
a = c sin , b = c cos
2
c c
a b
+
= 2
1 1
sin cos
+
= ( )
2
4 1 sin2
sin 2
+
= 2
1 14
sin 2 sin2
+
where 0 < <
2
2
min
c c
a b
+
= 8 when 2 = 90° = 45°
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(i) sin
Max. value = +1
Min value = -1
From 0ᵒ to 90ᵒ sin curve increasing
(ii) cos
Max value = +1
Min. value = -1
From 0ᵒ to 90ᵒ cos curve decreasing
(iii) tan
Min. value = - ∞
Max. value = +∞
From 0ᵒ to 90ᵒ tan curve increasing
(iv) cot
Min. value = - ∞
Max. value = + ∞
From 0ᵒ to 90ᵒ cot curve decreasing
(v) sec
Range (-∞, - 1] ⋃ [1, ∞)
(vi) cosec
Range (-∞, - 1] ⋃ [1, ∞)
♦ 0 < A, B < 90ᵒ
and A > B
then sinA > sinB
cosA < cosB
and tanA > tanB
------------------* * *----------------
Graphs of Trigonometric Ratios 10
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Maximum and minimum value of (a sin + b cos)
a sin + b cos
= 2 2
2 2 2 2
a ba b sin cos
a b a b
+ +
+ +
= 2 2a b+ [sin sin + cos cos]
= 2 2a b+ cos ( – ) ⇒– 1 cos ( – ) 1
( )2 2 2 2 2 2a b a b cos a b− + + − +
* Maximum value of sinn cosn is
n1
2
** * Inverse ratio
y = a tan2 x + b cot2 x Where a, b > 0
y= ( )2
a tanx b tanx 2 ab− +
as a tan x = b cot x tan2x = b
a (possible)
ymin = 2 ab
( ) y 2 ab,
Minimum value of a tan2 x + b cot2 x = 2 ab
* y = a cosec2 x + b sin2 x
Minimum value = 2 ab ⇒ ( ) y 2 ab,
* y = a sec2 x + b cos2 x
Minimum value = 2 ab
( ) y 2 ab,
* Minimum value of a sec2 + b cosec2
a sec2 + b cosec2
= a(1+tan2) + b(1+cot2)
= a + b + a tan2 + b cot2
Min value = a + b + 2 ab = ( )2
a b+
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1. why sin 1 is greater than sin1ᵒ?
Sol. sin1 = sin 57ᵒ.16’
As sin is increasing in 0,2
sin 57ᵒ 16’ > sin1ᵒ
2. which of the following is true?
(a) sin1 > sin1ᵒ (b) cos 1 < cos1ᵒ
(c) tan 1 > tan1ᵒ (d) all of these
Sol. (a) sin1 >sin1ᵒ above explained
(b) cos1 = cos 57ᵒ16’
as in 0,2
cos is decreasing
∴ cos57ᵒ16’ < cos1ᵒ
tan1 = tan57ᵒ16’
tan is increasing in 0,2
∴ tan 1 > tan1ᵒ
3 Which of the following is possible?
(a) sin = 4
3 (b) tan = 102
(c) sec = 3
4 (d) cos =
2
2
1
1
p
p
+
−(P≠±1)
Sol. sin = 4
3 is not possible as -1≤ sin≤1
tan = 102 is possible as tan can take any real value
sec =3
4 is not possible as sec ⋲ (-∞, -1]⋃[1, ∞)
cos = 2
2
1
1
p
p
+
− is not possible, as in
2
2
1
1
p
p
+
− numerator is
always greater than the denominator for any value of p
other than p = 0 hence 2
2
1
1
p
p
+
− does not lie in [-1, 1]
4 Find maximum and minimum value of sinx + cos x.
Sol. y = sin x + cos x
Max. value = 1 1 2+ =
Min. value = 1 1 2− + = −
5 y = 3 sin x + 4 cos x
Max. = 2 23 4+ = 5
Min. = 2 23 4− + = – 5
6 y = 5 sin x – 12 cos x + 40
Max. Value = 2 25 12+ + 10
= 13 + 10 = 23
Min. Value = 2 25 12− + + 10
= – 13 + 10 = – 3
7 Maximum value of (2 sin + 3 cos ) is.
(a) 2 (b) 13 (c) 15 (d) 1
Sol. (b)
y = 2 sin + 3 cos
max. value = 2 22 3 13+ + =
8 Find the minimum value of (24 sin + 7 cos ) is.
(a) –7 (b) 17 (c) –24 (d) - 25
Sol. (d)
y = 24 sin + 7 cos
min value = 2 224 7− + = - 25
9 Maximum and minimum values of 3 sin – 5 cos is.
(a) 3, –3 (b) 5, –5 (c) 34, 34− (d) 4, –4
Sol. (c)
y = 3 sin – 5 cos
max. value = 2 23 5 34+ + =
min. value = 2 23 5 34− + =
10 Find least value of 5 cos +9.
(a) 5 (b) 7 (c) –7 (d) –5
Sol. (b)
–1 cos 1 –5 5 cos 5
–5 + 12 5 cos + 12 5 + 12
7 5 cos + 12 17
Least value = 7
Or
Least value of cos = –1
Least value of 5 cos + 12 = – 5 + 12 = 7
11 Find least value of 4 sin – 3 cos + 5
Examples
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(a) 0 (b) 4 (c) 3 (d) 5
Sol. (a)
f() = 4 sin – 3 cos + 5
Least value of 4 sin – 3 cos = 2 24 3− + = – 5
Least value of f() = – 5 + 5 = 0
12 y = 8 sec2 x + 18 cos2 x
ymin = 2 8 18 = 2 × 12 = 24
13 The least value of a cosec2 + 16 sin2 is.
Sol. Least value = 2 9 16 = 2 × 12 = 24
14 The least value of 25 tan2 + 9 cot2 is.
Sol. Least value = 2 25 9 = 2 × 5 × 3 = 30
15 Minimum value of 49 sec2 + 9 cosec2 = ?
(a) 100 (b) 42 (c) 49 (d) 9
Sol. (a)
y = 49 sec2 + 9 cosec2
Minimum value = ( )2
49 9+ = (7 + 3)2 = 100
16 The least value of sin4 + cos4
(a) 1
4 (b) 3 (c)
1
2 (d)
3
4
Sol. (c)
sin4 + cos4 = 1 – 2 sin2 cos2 = 1 – 2sin 2
2
Value will be least when value of 2sin 2
2
is maximum.
max.
2sin 2
2 = ½, when = 450
Now max. value of sin4 + cos4 = − =
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1. Which is smaller, sin 67° or cos 67°?
(a) sin 67° (b) cos 67°
(c) Both are equal (d) None of these
2. If the sine of an angle is 2/5, then cosine of that angle is
(a) equal to 2/5 (b) less than 2/5
(c) greater than 2/5 (d) cannot be determined
3. The value of sincos ba + lies between
(a) ba − and ba + (b) a and b
(c) )( 22 ba +− and )( 22 ba + (d) 22 ba +− and 22 ba +
4. The maximum value of sin4cos3 − is
(a) 3 (b) 4 (c) 5 (d) None of these
5. The value of x, for maximum value of (sinx + cosx) is:
(a) 30° (b) 45° (c) 60° (d) 90°
6. Minimum value of 22 cos4sin5 + is
(a) 1 (b) 2 (c) 3 (d) 4
7. If 0 < < 2
, which of the following is true?
(a) sin2 + 2
1
sin < 2 (b) sin2 +
2
1
sin = 2
(c) sin2 + 2
1
sin > 2 (d) None of these
8. The maximum value of
+−
− xx
3cos
3cos 22
is
(a) 2
3− (b)
2
1 (c)
2
3 (d)
2
3
9. 22 cottan + is
(a) 2 (b) 2 (c) 2− (d) None of these
10. The value of x for the maximum value of
xx sincos3 + is
(a) 30° (b) 45° (c) 60° (d) 90°
11. The minimum value of 5sin4cos3 ++ xx is
(a) 5 (b) 9 (c) 7 (d) 0
12. The greatest and least value of xx cossin are
(a) 1,1 − (b) 2
1,
2
1− (c)
4
1,
4
1− (d) 2,2 −
13. The maximum value of xx 22 cos3sin4 + is
(a) 3 (b) 4 (c) 5 (d) 7
14. The maximum value of
++
+
6cos
6sin
xx in the
interval
2,0
is attained at
(a) 12
=x (b)
6
=x (c)
3
=x (d)
2
=x
15. The minimum value of 22 cot4tan9 + is
(a) 13 (b) 9 (c) 6 (d) 12
16. If ,sincos 42 +=A then for all values of
(a) 21 A (b) 116/13 A
(c) 16/134/3 A (d) 14/3 A
17. If ,cossin 42 +=A then for all real values of
(a) 21 A (b) 14
3 A
(c) 116
13 A (d)
16
13
4
3 A
18. What is the maximum value of sin6 + cos6 ?
(a) 1
2 (b)
1
4 (c) 1 (d) None of these
19. The value of x for maximum value of )cossin3( xx + is
(a) 30o (b) 45o (c) 60o (d) 90o
20. The least value of sin2 + cosec2 + cos2 + sec2 is:
(a) 3 (b) 4 (c) 5 (d) 6
21. The greatest value of 16sin x 8cos x is:
(a) 35 (b) 34 (c) 3 (d) 33
22. If O° < A < 90° and cos A – sin A > 0 then cos A + sin A
cannot be greater than:
(a) 1
3 (b)
1
2 (c)
1
2 (d) 2
Exercise - 1
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23. Let P = sin (sin + sin3 ). The P
(a) O only when O (b) O for all real
(c) O for all real (d) O only when O
24. The ratio of the greatest value of 2 – cos x + sin2 x to its
least value is:
(a) 1
4 (b)
9
4 (c)
13
4 (d)
7
4
25. The least value of sin2 + cos2 + tan2 + cot2 +
sec2 + cosec2 is:
(a) 8 (b) 7 (c) 5 (d) 6
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Answer key
1 b 2 c 3 d 4 c 5 b
6 d 7 c 8 c 9 a 10 a
11 d 12 b 13 b 14 a 15 d
16 d 17 b 18 b 19 c 20 c
21 a 22 d 23 c 24 c 25 b
1. (b)
cos 67° = cos (90° – 23°) = sin 23 °
In 1st quadrant as increases, the value of sin
increases.
sin 67° > sin 23°,
hence cos 67° is smaller.
2. (c)
cos = − = − = 2 4 21 41 sin 1
25 25 25
So, cos > 2
5
3. (d)
cos sina b + = 2 2a b+
++
+ 2222
sincos
ba
b
ba
a
= 2 2 sin( )a b + +
Since, ,1)sin(1– +
Then 2222 )sin( baba +++−
4. (c)
the maximum value of cossin ba + is 22 ba ++ and
minimum value is 22 ba +−
the maximum value is
(3cos 4sin ) + = 2 23 ( 4)+ − = 5
and the minimum value = - 5
5. (b)
sin x + cos x = 1 1
2 sinx cosx2 2
+
= 2 (sin x cos 45° + cos x sin 45°)
= 2 sin (x + 45°) = 2 , when x = 45°
Max. (sin x + cos x) = 2
6. (d)
Let y = 2 25sin 4cos + = 24 sin +
04)( + f )0sin( 2
The minimum value of )(f is 4
7. (c)
sin2 + 2
1
sin = sin2 +
2
1
sin – 2 + 2
=
21
sinsin
−
+ 2 > 2
0 0 sin 12
8. (c)
+−
− xx
3cos
3cos 22
cos cos3 3
x x
= − + + cos cos
3 3x x
− − +
= 2cos cos 2sin sin3 3
x x
= 2
sin sin 23
x
=3
sin 22
x
So maximum value is3
2 { 1 sin 2 1}x−
9. (a)
We know that2
10x
x
−
, 2
2
12 0x
x+ −
Put x = tan , 2
2
1tan 2
tan
+ 2 2tan cot 2 +
10. (a)
Let xxxf sincos3)( +=
+=
+=
3sin2sin
2
1cos
2
32)(
xxxxf
But 13
sin1
+−
x
Hence, )(xf is maximum, if 903
x
+ = 30x =
11. (d)
The minimum value of xx sin4cos3 + is 2 23 4− + = -
5
Solutions
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Hence the minimum value of 5sin4cos3 ++ xx
055 =+−=
12. (b)
Let y = sin cosx x = 1
sin 22
x
We know 1 sin2 1x− 1 1 1
sin 22 2 2
x−
Thus the greatest and least value of )(xf are 2
1 and
2
1− respectively
13. (b)
3sincos3sin4)( 222 ++= xxxxf and 1|sin|0 x
Maximum value of 3sin2 +x is 4
14. (a)
2 cos6 4
x
+ − = 2 cos
12x
−
Hence maximum value will be at 12
=x
15. (d)
the minimum value = 2 9 4
A.M. G.M.
2 29 tan 4cot
2
+ 2 24cot 9tan
2 2tan 4cot 12 +
Therefore, the minimum value is 12.
16. (d)
42 sincos +=A 222 sin.sincos +=A
22 sincos +A ]1sin[ 2
1A
Again 4242 sin)sin1(sincos +−=+=A
4
3
4
3
2
1sin
2
2 +
−= A
Hence, 14/3 A
17. (b)
We have 42 cossin +=A
22222 cossincoscossin ++=
(since )1cos2
11cossin 42 + A
Again, 4242 coscos1cossin +−=+
4
3
4
3
2
1cos1coscos
2
224 +
−=+−=
Hence, .14
3 A
18. (b)
x = sin6 + cos6 x = (sin2 )3 + (cos2 )3
x = (sin2 +cos2 )(sin4 + cos4 – sin2 cos2 )
x = 1 × [(sin2 + cos2 )2 – 3 sin2 cos2 ]
x = 1 – 3 sin2 cos2
x = 1 – 3
4 (2sin cos )2
x = 1 – 3
4 (sin 2)2 = 1 –
3
4sin 2
0 sin2 2 1
at sin2 2 = 0
x = 1 – 3
4 (0) = 1 and at sin2 2 = 1
x = 1 – 3
4 (1) =
1
4
i.e. 1
4 x 1
i.e. least value of sin6 + cos6 = 1
4
19. (c)
The greatest value of xx cossin3 + is 213 =+ and
obviously it will be at = 60x .
Alter :
+=
+
6sin2cos
2
1sin
2
32
xxx
As xsin is maximum at ,2
=x so
26
=+x or
3
=x
20. (c)
(sin2 + cos2 ) + (cosec2 + sec2 )
= 1 + (cosec2 + sec2 )
Minimum value of
(a cosec2 + b sec2 ) = ( )2
a b+
Minimum value of
(cosec2 + sec2 ) = ( )2
1 1+ = 4
min value = 4 + 1 = 5
21. (a)
16sin x8cos x = 24 sin x23 cos x = 24 sin x + 3 cos x
For maximum value,
4 sin x + 3 cos x must be maximum and maximum
value of:
4 sin x + 3 cos x = 2 24 3+ = 5
Greatest value of
16sin x8cos x = 25 = 32
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22. (d)
cos A – sin A > 0
sin A < cos A
tan A < 1 A < 45°
sin A + cos A < sin 45° + cos 45°
< +1 1
2 2
< 2
i.e. sin A + cos A cannot be greater than 2 .
23. (c)
P = sin (sin + sin3)
= sin (2 sin 2cos ) = sin 2 (2 sin .cos )
= sin 2 sin 2 = sin2 2 P = sin2 2 0
For all real
24. (c)
y = 2 – cos x + sin2 x = 2 – cos x + 1 – cos2 x
= – (cos2 x + cos x) + 3 =
21 1
cosx2 4
+ −
+ 3
=
213 1
cosx4 2
− +
Max. value of y occurs at cos x = 1
2− and it is
13
4
and min. value occurs at cos x = 1 and it is 1
The requited ratio is 13
4
25. (b)
sin2 + cos2 + tan2 + cot2 + sec2 + cosec2
= (sin2 + cos2 ) + tan2 + cot2 + (1 + tan2 ) + (1 +
cot2 )
= + 2(tan2 + cot2 )
Minimum value of
(a tan2 + b cot2 ) = 2 ab
Minimum value of
tan2 + cot2 = 2
min value = 3 + 2 × 2 = 7