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Revision Question Bank

Introduction to Trigonometry

1. Evaluate

2 0 2 2 0 2 2 0 0

02 0 2 0 2 0 2 0

sec 90 cot 2cos 60 tan 28 tan 62 cot 40

tan 502 sin 25 sin 65 3 sec 43 cot 47

Solution :

2

2 o 2 o

sec 90 cot

2 sin 25 sin 65

+

2 o 2 o 2 o o

o2 o 2 o

cos 60 tan 28 tan 62 cot 40

tan503 sec 43 cot 47

=

2 2

2 o 2 o o

cosec cot

2 sin 25 sin 90 43

+

2 2 o o

2 o 2 o o

12 tan 28tan 90 28

2

3 sec 43 cot 90 43

+

o

o o

cot 40

tan 90 40

osec 90 cosec

=

2 o 2 o2 2 o

o2 o 2 o2 2 o

12 tan 28 cot 28cosec cot cot 404

cot 403 sec 43 tan 432 sin cos 25

o o osin 90 cos ,tan 90 cot and cot 90 tan

=

111 2

2 1 3 3

2 2 2 2 2 2 1cosec cot 1,sin cos 1,sec tan 1 and tan 1

cot

= 1 1 3 1 6 10 5

12 6 6 6 3

2. Show that: 21 12sec

1 sin 1 sin

.

Solution :

LHS. =

1 1 1 sin 1 sin

1 sin 1 sin 1 sin 1 sin



= 2 2

2

2a b a b a b

1 sin

= 2

2

22sec

cos

2 2sin cos 1

LHS. = RHS.

3. Prove the identity (1 + cosec ) (1 – sin ) = cos cot .

Solution :

To prove, 1 cosec 1 sin =cos cot

LHS. = 1 cosec 1 sin

= 1 1

1 1 sin cosecsin sin

=

sin 1 1 sin

. 1 sinsin sin

= 2cos

sin

2 2sin cos 1

= cos

.cossin

= cos

cot .cos cotsin

LHS=RHS Hence proved

4. Without using trigonometric tables, evaluate:

2 0 2 0

2 0 2 0

cos 20 cos 70

sec 50 cot 40

× sec2 600 – 2 cot580cot 320 – 4 tan 130 tan 370 tan 450 tan 530 tan 770

Solution :

2 o 2 o2 o o o o o o o o

2 o 2 o

cos 20 cos 702sec 60 2cot58 cot32 4tan13 tan37 tan45 tan53 tan77

sec 50 cot 40

2 o 2 o o2 o o o o o

2 o 2 o o

o o o o

cos 20 cos 90 202 2 2cot58 cot 90 58 4tan13 tan37

sec 50 cot 90 50

1 tan 90 37 tan 90 13



= 2 o 2 o

o o

2 o 2 o

cos 20 sin 208 2cot58 tan58

sec 50 tan 50

o o o o4tan13 tan37 cot37 cot13

o ocos 90 sin ,ta 90 cot

= 1

8 2 1 4 1 1 tan cot 11

= 8 – 2 – 4 = 2

5. Prove the identity:1 cos A 1 cos A

4 cot A cosec .1 cos A 1 cos A

Solution :

To prove,

1 cosA 1 cosA4cot AcosecA

1 cosA 1 cosA

LHS = 1 cosA 1 cosA

1 cosA 1 cosA

=

2 21 cos A 1 cos A

1 cos A 1 cos A

=

2

1 cos A 1 cos A 1 cos A 1 cos A

1 cos A

2 2a b a b a b

= 2 2

2

2 2cosAsin A cos A 1

sin A

= 4cosA 1

. 4cot A.cosesAsin A sinA

= cos A 1

cot A ,cosecAsin A sin A

LHS =RHS Hence proved

6. Show that: (cosec A –1) (cosec A +1) (sec A –1) (sec A +1) =1.

Solution :

LHS



= cosecA 1 cosecA 1 secA 1 secA 1

= 1 1 1 1

1 1 1 1sin A sin A cos A cos A

1cosecA ,sec A

sin A cosA

= 1 sin A 1 sin A 1 cos A 1 cos A

sin A sin A cos A cos A

= 2 2

2 2

1 sin A 1 cos A.

sin A cos A

2 2a b a b a b

=2 2

2 2

2 2

cos A sin A.cos sin A cos A 1

sin A sin A

= 1

LHS=RHS

7. Show that: 2

2

2 cosec A sin A cosA

cosec A 2cot A sin A cosA

.

Solution :

LHS = 2 2

2

2

12

2 cosec A sin A1 2cos Acosec A 2cot A

sin A sin A

1 cosAcosecA and cot A

sin A sin A

=

2

2

2

2sin A 1

sin A1 2sin Acos A

sin A

= 2 2

2

2sin A 1 sin A

Sin A 1 2sin Acos A

= 2

2 2

2sin A 1

sin A cos A 2sin Acos A

2 2sin A cos A 1



=

2 2 22sin A sin A cos A

sin A cos A

2 21 sin cos

=

2 2 2

2

2sin A sin A cos A

sin A cos A

=

2 2

2

sin A cos A

sin A cos A

=

2

sin A cos A sin A cos A

sin A cos A

2 2a b a b a b

= sin A cos A

sin A cos A

LHS=RHS

8. If 3 cot A = 4 , then check whether

22 2

2

1 tan Acos A sin A or not.

1 tan A

Solution :

Let us consider a right angled ABC in which oB 90 . For A ,

Base = AB and perpendicular = BC. Also, hypotenuse = AC

3cot A 4

4cot A

3 ……..(i)

But cot A = Base AB

perpendicular BC

From(i) and (ii) we get

AB 4 4k

BC 3 3k

AB = 4k and BC = 3k

Using Pythagoras theorem

AC2 = AB2 + BC2

AC2 = (4k)2 + (3k)2



AC = 2 216k 9k

= 2225k 5k 5k

Now, sin A= BC 3k 3

AC 5k 5

cos A = AB 4k 4

AC 5k 5

Also, tan A = BC 3k 3

AB 4k 4

Now, to check the given equation,

2

2

22

3 91 11 tan A 4 16LHS

91 tan A 3 11 164

=

16 9 7716 16

16 9 25 2516 16

RHS = 2 2cos A sin A

= 2 2

4 3 16 9 16 9 7

5 5 25 25 25 25

From Eqs. (i) and (ii),

LHS=RHS

2

2 2

2

1 tan Acos A sin A

1 tan A

9. Prove the identity (cot A – tan A) cos A = cosec A – 2 sin A.

Solution :

To prove (cot A – tan A) cos A = cosec A – 2sin A

LHS = (cot A – tan A) cos A

= cos A sin A

cos Asin A cos A



= 2 2cos A sin A

.cos Asin Acos A

= 2 2cos A sin A

sin A

= 2cos A

sin Asin A

= 21 sin A

sin Asin A

= 2 21sin A sin A cos 1 sin

sin A

= cosec A – 2 sin A

LHS = RHS Hence proved

10. Show that cos 1 1

1 sin 1 sin

can be written in the form k tan and find the

value of k.

Solution :

1 1cos

1 sin 1 sin

= 1 sin 1 sin

cos1 sin 1 sin

= 2

2sincos

1 sin

2 2a b a b a b

= 2

2sin sincos . 2

cos cos

= 2tan

k 2 sin

tancos



Chapter Test {Trigonometry}

M: Marks: 40 M: Time: 40 Min.

1. Without using trigonometrical tables, evaluate 0 0 0

0 0 0 0 0 0 0

cos58 sin 22 cos 38

sin32 cos 68 tan18 tan35 tan60 tan72 tan55

[4]

Solution: 0 0

0 0

cos 58 sin 22

sin 32 cos 68

0

0 0 0 0 0

cos 38

tan 18 tan35 tan 60 tan 72 tan55

= 0 0 0 0

0 0

cos 90 32 sin 90 68

sin 32 cos 68

0 0 0

0 0 0 0 0 0

cos 38 cosec 90 38

tan18 tan35 3 tan 90 38 tan 90 35

= 0 0

0 0

sin 32 cos 68

sin 32 cos 68 –

0 0

0 0 0 0

cos 38 sec 38

tan 18 tan 35 3 cot 18 cot35

0 0

0

0

cos 90 sin , sin 90 cos ,

cosec 90 sec

and tan 90 cot

= 0 0 0 0

11 1

tan 18 cot18 tan 35 cot 35 3

[ cos . sec 1 ]

= 1

1 11.1. 3

= 2 3 1

3



2. If 4 4sin cos 1

a b a b

, then prove that

8 8

33 3

sin cos 1

a b a b

. [4]

Solution:

To prove

8 8

33 3

sin cos 1

a b a b

Given, 4 4sin cos 1

a b a b

4 4

2sin cosa b 1

a b

4 4

22 2sin cos

a b sin cosa b

[ 2 2sin cos 1 ]

4 4a b a bsin cos

a b

= 4 4 4 2sin cos sin cos

Þ 4 4 4 4b asin sin cos cos

a b

= 4 4 2 2sin cos 2sin cos

4 4 2 2b asin cos 2 sin cos 0

a b

2 2

2 2b asin cos

a a

– 2 2b a

2 sin cos 0a a

2

2 2b asin cos 0

a b

[

22 2a b 2ab a b ]

2 2b asin cos 0

a b [taking square root]

2

2

asin abcos bb

a

2 2sin cos

k saya b

..(i)

2 2sin ak and cos bk



2 2sin cos ak bk

ka b a b

..(ii)

From Eqs. (i) and (ii), we get 2 2 2 2sin cos sin cos

a b a b

2 2sin cos 1

a b a b

Taking first and third terms, 2

2sin 1 asin

a a b a b

.(iii)

Taking second and third terms, 2cos 1

b a b

2 bcos

a b

..(iv)

Now,

4 42 28 8

3 3 3 3

sin cossin cos

a b a b

=

4 4

3 3

a ab

a b a b

a b

[ from Eqs (iii) and (iv)]

=

4 4

a b

a b a b

=

4 3

a b 1

a b a b

3. If 7 2 2sin 3cos 4 and is an acute angle, then prove that:2

sec cosec 2 .3

[4]

Solution:

Given, 2 27sin 3cos 4

2 2 24sin 3sin 3 cos 4

2 2 24sin 3 sin cos 4

24sin 3 4 [ 2 2sin cos 1 ]

24sin 4 3

24sin 1



2 1sin

4

1

sin2

[taking positive square root as is acute angle]

cosec 2 1

sincosec

and 2cos 1 sin

= 2

11

2

= 1 4 1

14 4

= 3 3

4 2

2

sec3

1

seccos

2sec cosec 2

3

Hence proved.

4. Find the acute angles of A and B, if sin (A + 2B) = 3

2 and cos (A + 4B) = 00, where A > B.

[4]

Solution:

Given that, sin (A + 2B) = 3

2

sin(A + 2B) = sin600

A+ 2B = 60° ...(i)

and cos(A + 4B) = 00

cos (A + 4B) = cos 90°

A+4B = 90° ...(ii)

On subtracting Eq. (i) from Eq. (ii), we get



0

0

0

A 4B 90

A 2B 60

2B 30

B = 15°

On putting B = 15° in Eq. (i) we get

A + 2(15°) = 60°

A =60° – 30° = 30°

5. If (cot + tan ) = m and (sec – cos ) = n, then prove that: 2/3 2/3

mn mn 1 . [4]

Solution:

To prove, (m n)2/3 – (mn )2/3 =1

Given, (cot + tan ) = m and (sec – cos ) = n

Now, m2n= (cot tan )2 (sec cos )

=

2cos sin 1

cossin cos cos

=

2 2cos 1 cos

sin cos cos

= 2

2 2

2 2

1 sin. [ cos sin 1]

sin cos cos

2 3

3

1m n sec

cos

..(i)

and 22mn cot tan sec cos

=

2cos sin 1

cossin cos cos

=

22 2 2cos sin 1 cos

sin cos cos

=

22

2

sin1

sin cos cos

[ 2 2sin cos 1 ]

= 4 3

2 3

3 3

sin sinmn tan

sin cos cos

…(ii)

From Eq. (i), sec3 = m2n

sec = (m2n)1/3 [taking cube root both sides] ..(iii)



From Eq. (ii), tan3 = mn2

sec = (m2n )1/3 [taking cube root both sides] ...(iv)

On squaring Eqs. (iii) and (iv) and then subtracting, we get

sec+ – tan2 = (m2n)2/3 – (mn2)2/3

1 = (m2n)2/3 –(mn2)2/3 [ sec2 – tan2 = 1]

Hence proved.

6. Prove that :tan cot

1 cot tan1 cot 1 tan

. [4]

Solution:

To prove, tan cot

1 cot 1 tan

=

1

tan tan1 cot tan

1 1 tan1tan

LHS = tan cot

1 cot 1 tan

=

1

tan tan1 1 tan1

tan

1

cottan

=

tan 1

tan 1 tan 1 tantan

=

2 3tan 1 tan 1

tan 1 tan tan 1 tan tan 1

=

2tan 1 tan tan 1

tan tan 1

[ a3 – b3 = (a – b) (a2 + ab + b2)]

= 2 2tan tan 1 tan tan 1

tan tan tan tan

= tan 1 cot 1 cot tan

LHS = RHS Hence proved.



7. Prove that :1 1 1 1

.cosec cot sin sin cosec cot

[4]

Solution:

To prove,

1 1 1 1

cosec cot sin sin cosec cot

LHS = 1 1

cosec cot sin

=

cosec cot1cosec

cosec cot cosec cot

1cosec

sin

=

2 2

cosec cotcosec

cosec cot

2 2a b a b a b

= cosec cot

cosec1

[ 2 2cosec cot 1 ]

= cosec cot cosec

= cot

RHS = 1 1

sin cosec cot

=

cosec cot1cosec

cosec cot cosec cot

=

2 2

cosec cotcosec

cosec cot

= cosec cot

cosec1

[ 2 2cosec cot 1 ]

= cosec cosec cot

= cot


8. If 1 1

cos and tan2 3

then find, Sin where and are both acute angles. [4]

Solution:

Here, cos 1

2 cos 60° 0 1

cos 602

a =60°



and 1

tan3

= tan 30° 0 1tan 30

3

030

0 0sin sin 60 30

= sin 90° = 1 [ sin 90° = 1]

Hence, sin ( ) is 1.

9. Without using trigonometric tables evaluate the following

2 0 2 2 0 2 0 2 0

2 0 2 0 2 0 2 0

cosec 90 tan 2tan 30 sec 52 sin 38

4 cos 48 cos 42 cosec 70 tan 20

[4]

Solution:

Now,

2 0 2 2 0 2 0 2 0

2 0 2 0 2 0 2 0

cosec 90 tan 2tan 30 sec 52 sin 38

4 cos 48 cos 42 cosec 70 tan 20

=

0 2 2

0 0 2 2 0

[{cosec 90 } tan ]

4[{cos 90 42 } cos 42 ]

2

0 0 2 2 0

2 0 0 2 2 0

12 . [sec 90 38 } . sin 38

3

{cosec 90 20 } tan 20

=

2 0 2 02 2

2 0 2 0 2 0 2 0

2.cosec 38 . sin 38

sec tan 34{sin 42 cos 42 } sec 20 tan 20

0

0

0

cosec 90 sec

cos 90 sec

and sec 90 cosec

= 2 0

2 0

1 2 1. . sin 38

4 3 sin 38

2 2

2 2

sec tan 1

sin cos 1

1and cosec

sin



= 1 2 1

.4 3 1

= 3 8 5

12 12

10. Prove that:

3 3

1 cot A tan A sin A cos A

sec A cosec A

= sin2A cos2A. [4]

Solution:

To prove,

2 2

3 3

1 cot tan A sin A cos Asin Acos A

sec A cosec A

LHS =

3 3

1 cot A tan A sin A cos A

sec A cosec A

=

3 3

cos A sin A1 sin A cos A

sin A cos A

1 1

cos A sin A

sin A cos Atan A , cot A

cos A sin A

1 1and sec A , cosec A

cos A sin A

=

2 2

3 3

3 3

cos A sin A1 sin A cos A

sin A cos A

sin A cos A

sin A cos A

=

3 3

3 3

11 sin A cos A

sin A cos A

sin A cos A

sin Acos A

3 3

2 2

3 3

sin Acos A sin A cos Asin A cosA 1[sin cos 1]


= sin A cos A 1

sin A cos A

×

3 3

2 2


sin A cos A sin A cos A sin Acos A

[ a3 – b3 = (a – b) (a2 + b2 + ab)]



= 2 2

2 2

sin A cos A 1 sin Acos A


= 2 2sin A cos A 1 sin Acos A

1 sin A cos A

[ sin2A + cos2 A = 1]

= sin2 A cos2 A


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