Question Bank with Solutions Inverse Trigonometric Functions

INVERSE TRIGONOMETRIC FUNCTIONS

One Mark Questions (MCQ) 13. One branch of cos −1 𝑥 other than the principal
1. The principal branch of sin−1 𝑥 is value branch corresponds to
𝜋 3𝜋 3𝜋
𝜋 𝜋 𝜋 𝜋 a) [ 2 , ] b) [𝜋, 2𝜋] − { 2 }
a) [− 2 , 2 ] b) (− 2 , 2 ) 2

c) [0, 𝜋] d) (0, 𝜋) c) (0, 𝜋) d) [2𝜋, 3𝜋]

𝜋 𝜋
2. The principal branch of cos −1 𝑥 is 14. The range of 𝑓(𝑥) = sin−1 𝑥 other than [− , ] is
2 2
𝜋 𝜋 𝜋 𝜋 𝜋 𝜋 3𝜋
a) [− 2 , 2 ] b) (− 2 , 2 ) a) [− 2 , 𝜋] b) [ 2 , ]
2
c) [0, 𝜋] d) (0, 𝜋) c) [0 , 𝜋] d) [−𝜋, 𝜋]
3. The principal branch of tan−1 𝑥 is 15. The principal value of cos −1 ( 2 ) is
√3
𝜋 𝜋 𝜋 𝜋
a) [− 2 , 2 ] b) (− 2 , 2 ) 𝜋 𝜋 𝜋 𝜋
a) 3 b) 6 c) 2 d) 4
c) [0, 𝜋] d) (0, 𝜋) 1
16. The principal value of sin−1 (− 2) =
4. The principal branch of cot −1 𝑥 is
𝜋 𝜋 𝜋 𝜋
𝜋 𝜋 𝜋 𝜋 a) 3 b) − 3 c) 6 d) − 6
a) [− 2 , 2 ] b) (− 2 , 2 )
√3
c) [0, 𝜋] d) (0, 𝜋) 17. The principal value of cos −1 (− 2
) =
5. The principal branch of sec −1 𝑥 is a) 3
𝜋
b) 6
𝜋
c)
5𝜋
d)
2𝜋
𝜋 𝜋 𝜋 6 3
a) [0, 𝜋] − { 2 } b) [− 2 , 2 ] − {0} 18. Principal value of tan−1(−1)
𝜋 𝜋 𝜋 𝜋 𝜋 3𝜋 5𝜋
c) (− 2 , 2 ) − {0} d) (0, 𝜋) − { 2 } a) 4 b) − 4 c) d)
4 4
6. The principal branch of cosec −1 𝑥 is 19. The principal value of cot −1(−√3) =
𝜋 𝜋 𝜋 𝜋 2𝜋 5𝜋 𝜋
a) [0, 𝜋] − { 2 } b) [− 2 , 2 ] − {0} a) b) c) d)
3 3 6 4
𝜋 𝜋 𝜋
c) (− 2 , 2 ) − {0} d) (0, 𝜋) − { 2 } 20. The domain of 𝑓(𝑥) = cos −1 (2𝑥 − 1) is
7. The domain of 𝑓(𝑥) = sin−1 𝑥 is a) (0, 1) b) [0, 1]
a) (−1, 1) b) [−1, 1] c) (0, 1] d) [0, 1)
−1
c) (0, 1] d) 𝑅 21. The domain of sin 2𝑥 𝑖𝑠
1 1
8. −1
The domain of 𝑓(𝑥) = cos 𝑥 is a) [0,1] b) [−1, 1] c) [− , ] d) [−2, 2]
2 2
a) (−1, 1) b) [−1, 1] 22. The principal value of cos −1 (cos
3𝜋
)+ sin−1 sin 4
𝜋
4
c) (0, 1] d) 𝑅
−1
is
9. The domain of 𝑓(𝑥) = tan 𝑥 is 7𝜋 𝜋 𝜋
a) 𝜋 b) c) − 3 d) 2
a) 𝑅 b) 𝑅 − [−1, 1] 6
1
c) 𝑅 − (−1, 1) d) [−1, 1] 23. tan−1 (2 cos (2 sin−1 (2))) =
10. The domain of 𝑓(𝑥) = cot −1 𝑥 is a) 3
𝜋
b) 6
𝜋
c) 4
𝜋
d) 0
a) (−1, 1) b) 𝑅 − [−1, 1] 1
24. The value of sec −1 √2 − cos−1 ( 2) is
c) 𝑅 − (−1, 1) d) (−∞, ∞) √

11. The domain of 𝑓(𝑥) = sec −1 𝑥 is a) 0 b) 1 c) 2 d) √2

a) (−∞, −1) ∪ (1, ∞) b) (−∞, −1) ∪ [1, ∞) 25. tan−1 √3 − cot −1(−√3) is equal to
𝜋
c) (−∞, −1) ∩ (1, ∞) d) (−∞, −1] ∪ [1, ∞) a) 𝜋 b) − 2 c) 0 d) 2√3
12. The domain of 𝑓(𝑥) = cosec −1 𝑥 is 26. The value of [cos−1(−1) − sin−1(1)] is
a) 𝑅 − (−1, 1) b) 𝑅 − [−1, 1] 𝜋 3𝜋 3𝜋
a) 2 b) 𝜋 c) − 2
d) 2
c) (−1, 1] d) [−1, 1)
27. tan−1 √3 − sec −1 (−2)is equal to
𝜋 𝜋 2𝜋
a) 𝜋 b) − 3 c) 3 d) 3

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Question Bank with Solutions Inverse Trigonometric Functions

13𝜋 1
28. cos−1 (𝑐𝑜𝑠 6
) is equal to 8. The value of cos (2 sin−1 (2)) is ______
13𝜋 𝜋 13𝜋 7𝜋
a) b) 6 c) − d) 9. The value of sin[𝑡𝑎𝑛−1 √3 − cot −1(−√3)] is _____
6 6 6
29. sin(tan−1 𝑥) , |𝑥| < 1 is equal to 10. The cos[tan−1 √3 − sec −1 (−2)] is ______
𝑥 1 𝜋 √3
a)
√1−𝑥 2
b)
√1−𝑥 2
11. The value of sin [ − sin−1 (− )] is ______
3 2
1 𝑥 𝜋 1
c)
√1+𝑥 2
d)
√1+𝑥 2
12. The value of cos [ 3 − sin−1 (− 2)] is ______
−1 (−𝑥) −1
30. The value of 𝑥 for which sin = − sin 𝑥 13.
𝜋 1
The value of sin [ 3 − sin−1 (− 2)] is ______
holds is 𝜋 1
𝜋 𝜋 14. The value of sin [ 2 − sin−1 (− 2)] is ______
a) [0, 𝜋] b) [− 2 , 2 ] c) [−1,1] d) [0,1]
7𝜋
31. The set of value of 𝑥 for which 15. The principal value of cos −1 (cos 6 ) is ______
3𝜋
3 sin−1 𝑥 = sin−1(3𝑥 − 4𝑥 3 ), holds 16. The principal value of cos −1 (cos 2 ) is ______
1 1 √3 √3 3𝜋
a) [− 2 , 2] b) [− , ]
2 2 17. The value of sin−1 (sin ( 5 )) is ______
1 1 √3 √3 2𝜋
c) [− 2 , 2) d) (− 2 , 2 ] 18. The value of sin−1 (sin ( )) is ______
3
32. The set of value of 𝑥 for which −1 3𝜋
19. The value of tan (tan 4 ) is ______
3 cos −1 𝑥 = cos−1(4𝑥 3 − 3𝑥), holds 8
1 1 1 1 20. The value of tan (cos −1 17) is ______
a) [2 , 1] b) (2 , 1) c) [2 , 1) d) (2 , 1]
1
21. The value of cos[sin−1 3 + sec −1 3] is ______
33. The set of value of 𝑥 for which
1
3 tan−1 𝑥 = tan−1 (
3𝑥−𝑥 3
), holds 22. If sin [sin−1 5 + cos −1 𝑥] = 1, then 𝑥 is______
1−3𝑥 2
−1 −1
1 1 1 1 23. The value of cot(tan 𝑥 + cot 𝑥) =______
a) [− , ] b) (− , )
√3 √3 √3 √3
1 1 1 1
c) [− 3 , 3) d) (− 3 , 3]
√ √ √ √
34. The set of value of 𝑥 for which ANSWER KEY
sin−1(2𝑥√1 − 𝑥 2 = 2 sin−1 𝑥, holds MCQ
1 1 1 1 QN KEY QN KEY QN KEY QN KEY QN KEY
a) [− , ] b) [− , ]
√2 √2 2 2
1 √3 √3
1 a 2 c 3 b 4 d 5 a
c) [2 , 1] d) [− , ]
2 2 6 b 7 b 8 b 9 a 10 d
35. The set of value of 𝑥 for which
11 d 12 a 13 d 14 b 15 b
sin−1(2𝑥√1 − 𝑥 2 ) = 2 cos−1 𝑥, holds
16 d 17 c 18 b 19 c 20 b
1 1 1
a) [ 2 , 1] b) [− 2 , 2] 21 c 22 a 23 c 24 a 25 b
√
1 √3 √3 26 a 27 b 28 b 29 d 30 c
c) [2 , 1] d) [− 2 , 2 ]
31 a 32 a 33 b 34 a 35 a

One Mark Questions (Fill in the blanks) FILL IN THE BLANK

1 QN KEY QN KEY QN KEY QN KEY QN KEY
1. The principal value of 𝑐𝑜𝑠 −1 (2) is ______ 𝜋 𝜋
2𝜋 2𝜋
1 1
3
2 3 𝜋 4 5 −3
3 3
2. The principal value of cos −1 (− ) is ______
2 2𝜋 1 1 1
−1 6 7 8 9 −1 10
3. The principal value of cos (−1) is ______ 3 2 2 2

1 √3 √3 5𝜋
4. The principal value of cot −1 (− ) is ______ 11 12 0 13 1 14 15
2 2 6
√3
𝜋 2𝜋 𝜋 𝜋 15
5. The principal value of 𝑡𝑎𝑛−1 (−√3) is ______ 16 17 18 19 − 20
2 5 3 4 8
1 1 1
6. The value of cos −1 (2) + 2𝑠𝑖𝑛−1 (2) is ______ 21 0 22 5
23 0
−1
7. The value of sin[𝑐𝑜𝑠𝑒𝑐 (2)] is ______

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Two or Three Marks Questions 6) 𝐭𝐚𝐧−𝟏 √𝟑 − 𝐬𝐞𝐜 −𝟏 (−𝟐)

1) Find the value of the following Solution: tan−1 √3 − sec −1(−2)
𝝅 𝟏 𝜋 𝜋
1) 𝐬𝐢𝐧 [ 𝟑 − 𝐬𝐢𝐧−𝟏 (− 𝟐)] = tan−1 √3 − (𝜋 − sec −1 2) = − 𝜋 +
3 3
Solution: 𝜋−3𝜋+𝜋 𝜋
= 3
= −3
𝜋 1 𝜋 1
sin ( 3 − sin−1 (− 2)) = sin (3 − (− sin−1 2))
𝟏
𝜋 𝜋
= sin ( 3 + 6 ) = sin (
2𝜋+𝜋
) 7) 𝐭𝐚𝐧−𝟏 (𝟐 𝐜𝐨𝐬 (𝟐 𝐬𝐢𝐧−𝟏 ))
6 𝟐
1
3𝜋
= sin ( 6 ) = sin 2 = 1
𝜋 Solution: tan−1 (2 cos (2 sin−1 2))
𝜋 𝜋
= tan−1 [2 cos (2 6 )] = tan−1 [2 cos 3 ]
𝟐
2) 𝐬𝐢𝐧−𝟏 (𝐬𝐢𝐧 𝟑 ) 1
= tan−1 [2 ] = tan−1 1 =
𝜋
2 4
2𝜋
Solution: sin−1 (sin 3
)= sin−1 (sin 1200 )
𝟏 𝟏
= sin−1 sin(1800 − 60 0)
8) 𝐜𝐨𝐬 −𝟏 (𝟐) + 𝟐 𝐬𝐢𝐧−𝟏 (𝟐)
𝜋
= sin−1 sin 600 = 600 = 3 1 1
Solution: cos−1 (2) + 2 sin−1 (2)
𝜋 𝜋 𝜋 𝜋 2𝜋
−𝟏 𝟑𝝅 =3 +26 = 3 + 3 = 3
3) 𝐬𝐢𝐧 (𝐬𝐢𝐧 )
𝟓
3
Solution: sin−1 (sin ) = sin−1 𝑠𝑖𝑛 (1080 ) 𝟏 𝟏
5 9) 𝐭𝐚𝐧−𝟏(𝟏) + 𝐜𝐨𝐬 −𝟏 (− 𝟐) + 𝐬𝐢𝐧−𝟏 (− 𝟐)
−1 0 0) −1 0
= 𝑠𝑖𝑛 sin(180 − 72 = 𝑠𝑖𝑛 sin 72 1 1
3 Solution: tan−1 (1) + cos −1 (− 2) + sin−1 (− 2)
⟹ sin−1 (sin ) = 72 0
5 𝜋 1 1
𝑶𝑹 = 4 + 𝜋 − cos−1 (2) − sin−1 (2)
3 2𝜋 𝜋 𝜋 𝜋 3𝜋+12𝜋−4𝜋−2𝜋 9𝜋 3𝜋
sin−1 (sin ) = sin−1 𝑠𝑖𝑛 (𝜋 − ) = 4+𝜋−3−6 = 12
= 12 = 4
5 5
2𝜋 2𝜋
= 𝑠𝑖𝑛−1 sin ( 5 ) = 5
2) Prove the following
−𝟏 𝟏
𝟏𝟑𝛑 1) 𝐬𝐢𝐧−𝟏 (𝟐𝐱√𝟏 − 𝐱 𝟐 ) = 𝟐 𝐬𝐢𝐧−𝟏 𝐱 , ≤𝐱≤ .
4) 𝐜𝐨𝐬 −𝟏 (𝐜𝐨𝐬 ) √𝟐 √𝟐
𝟔 −1
13𝜋 𝜋
Solution: 𝐋et x = sinθ ⇒ θ = sin x
Solution: cos−1 cos ( ) = cos −1 cos (2𝜋 + 6 )
6 LHS = sin−1 (2x√1 − x 2 )
−1𝜋 𝜋
= cos cos 6 = 6 𝑶𝑹
= sin−1(2sinθ√1 − sin2 θ)
13𝜋
cos−1 cos ( 6 ) = cos −1 cos(3900 ) = sin−1(2sinθ√cos 2 θ)
−1
= cos cos(360 + 300 ) = cos −1 cos(300 )
0
= sin−1(2sinθ. cosθ) = sin−1 (sin2θ)
𝜋
= 300 = = 2θ = 2 sin−1 x = RHS
6

𝟕𝛑 𝟏−𝐱 𝟐
5) 𝐜𝐨𝐬 −𝟏 (𝐜𝐨𝐬 ) 2) 𝟐 𝐭𝐚𝐧−𝟏 𝐱 = 𝐜𝐨𝐬 −𝟏 (𝟏+𝐱 𝟐) , 𝐱 ≥ 𝟎
𝟔

Solution: cos−1 (cos

7𝜋
) = cos−1(cos 2100 ) Solution: Let 𝑥 = tan 𝜃 ⟹ 𝜃 = tan−1 𝑥
6
1−𝑥 2 1−𝑡𝑎𝑛2 𝜃
= cos−1(cos(1800 + 30)) = cos−1 (−𝑐𝑜𝑠 300 ) RHS = cos −1 (1+𝑥2 ) = cos−1 (1+𝑡𝑎𝑛2 𝜃)
= 1800 − cos−1(𝑐𝑜𝑠 300 ) = 1800 − 300 = cos −1 cos 2𝜃 = 2𝜃 = 2 tan−1 𝑥 = LHS.
7𝜋
⟹ cos−1 (cos ) = 1500
6
𝟏 𝟏
𝑶𝑹 3) 𝟑 𝐬𝐢𝐧−𝟏 𝐱 = 𝐬𝐢𝐧−𝟏 (𝟑𝐱 − 𝟒𝐱 𝟑 ), 𝐱 ∈ [− 𝟐 , 𝟐]
7𝜋 5𝜋
cos−1 (cos ) = cos −1 (cos (2𝜋 − )) Solution: Let x = sinθ ⇒ θ = sin−1 x
6 6
5𝜋 5𝜋 RHS = sin−1(3x − 4x 3 )
= cos−1 (cos ) =
6 6 = sin−1 (3sinθ − 4 sin3 θ)
= sin−1(sin3θ) = 3θ = 3 sin−1 x = LHS

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𝟏 √sec2 θ−1 secθ−1

4) 𝟑 𝐜𝐨𝐬 −𝟏 𝐱 = 𝐜𝐨𝐬 −𝟏 (𝟒𝐱 𝟑 − 𝟑𝐱) 𝐱 ∈ [𝟐 , 𝟏] . = tan−1 ( )= tan−1 ( )
tanθ tanθ
Solution: Let x = cosθ ⇒ θ = cos−1 x 1
−1
1−cosθ

RHS = cos −1 (4x 3 − 3x) = tan−1 (sinθ/cosθ

cosθ
) = tan−1 ( cosθ
sinθ )
cosθ
= cos−1(4 cos3 θ − 3cosθ) 1−cosθ sinθ/2
= tan−1 ( ) = tan−1 (cosθ/2)
= cos −1 (cos3θ) = 3θ = 3 cos−1 𝑥 = 𝐿𝐻𝑆 sinθ
θ tan−1 x
= tan−1 (tanθ/2) = =
2 2
√𝟏+𝐬𝐢𝐧 𝐱 + √𝟏−𝐬𝐢𝐧𝐱 𝐱
8) 𝐜𝐨𝐭 −𝟏 ( ) =𝟐
√𝟏+𝐬𝐢𝐧 𝐱− √𝟏−𝐬𝐢𝐧 𝐱
√1+sin 𝑥 + √1−sin 𝑥 𝒄𝒐𝒔 𝒙
Solution: LHS = cot −1 [ ] 5) 𝐭𝐚𝐧−𝟏 (𝟏−𝒔𝒊𝒏 𝒙)
√1+sin 𝑥 – √1−sin 𝑥
𝜋
𝑥 𝑥 𝑥
cos + sin + cos – sin
𝑥 cos 𝑥 sin ( −𝑥)
= cot −1
[ 2 2 2 2
] Solution: tan−1 ( ) = tan−1 ( 2
𝜋 )
𝑥 𝑥 𝑥 𝑥 1−sin 𝑥 1−cos( − 𝑥)
cos + sin −( cos − sin ) 2
2 2 2 2
𝑥 𝑥 𝑥 𝜋⁄ −𝑥 𝜋⁄ −𝑥
cos + cos 2 cos 2 sin( 2 ) cos( 2 )
−1 2 2 −1 2 −1 22
= cot [ 𝑥 𝑥 𝑥 𝑥 ] = cot [ 𝑥 ] = tan [ 𝜋⁄ −𝑥 ]
cos + sin − cos + sin 2 sin 2 2
2 2 2 2 2 1−[1−2 𝑠𝑖𝑛 ( )]
2
𝑥 𝑥
= cot −1 cot 2 = 2 = 𝑅𝐻𝑆 𝜋
2 sin( − ) cos( − )
𝑥 𝜋 𝑥 𝜋 𝑥
cos( − )
−1 4 2 4 2 −1 4 2
= tan [ 𝜋 𝑥 ] = tan [ 𝜋 𝑥 ]
2𝑠𝑖𝑛2 ( − ) sin( − )
4 2 4 2
3) Write the following in simplest form 𝜋 𝑥 𝜋 𝜋 𝑥
= tan−1 [cot ( − )] = − cot −1 cot ( − )
𝐜𝐨𝐬 𝐱−𝐬𝐢𝐧 𝐱 4 2 2 4 2
1) 𝐭𝐚𝐧−𝟏 ( 𝐜𝐨𝐬 𝐱+𝐬𝐢𝐧 𝐱 ) , 𝟎 <𝐱<𝛑 𝜋 𝜋 𝑥 𝜋 𝜋 𝑥 2𝜋−𝜋 𝑥 𝜋 𝑥
= −[ − ]= − + = + = +
cos x sin x
− 2 4 2 2 4 2 4 2 4 2
cos x−sin x
Solution: tan−1 ( ) = tan−1 ( cos x cos x
cos x sin x )
cos x+sin x +
cos x cos x
𝟏 𝟐𝒙 𝟏 𝟐
−𝟏 𝟏−𝒚
=
1−tan x
tan−1 (1+tan x) = tan −1 π
(tan ( 4 − x)) 6) 𝐭𝐚𝐧 [ 𝐬𝐢𝐧−𝟏 ( ) + 𝐜𝐨𝐬 ( )]
𝟐 𝟏+𝒙𝟐 𝟐 𝟏+𝒚𝟐
π −1
= −x Solution: Let 𝑥 = tan 𝛼 ⟹ 𝛼 = tan 𝑥
4 −1
𝑦 = tan 𝛽 ⟹ 𝛽 = tan 𝑦
1 2𝑥 1 1−𝑦 2
𝟏−𝐜𝐨𝐬 𝒙 Now, tan [2 sin−1 (1+𝑥2 ) + 2 cos −1 (1+𝑦2 )]
2)𝐭𝐚𝐧−𝟏 (√𝟏+𝐜𝐨𝐬 𝒙) 𝒘𝒉𝒆𝒓𝒆 𝟎 < 𝒙 < 𝝅
1 2 tan 𝛼 1 1−𝑡𝑎𝑛2 𝛽
= tan [2 sin−1 (1+𝑡𝑎𝑛2 𝛼) + 2 cos−1 (1+𝑡𝑎𝑛2 𝛽)]
√1−cos x 2 sin2 x/2
Solution: tan−1 ( ) = tan−1 (√ ) 1 1
√1+cos x 2 cos2 x/2 = tan [2 sin−1 sin 2𝛼 + 2 cos−1 cos 2𝛽]
x 1 1
sin2 x = tan [2 2𝛼 + 2 2𝛽] = tan[𝛼 + 𝛽]
= tan−1 (√ 2
x ) = tan−1 (√tan2 2)
cos2 tan 𝛼+tan 𝛽 𝑥+𝑦
2
= 1−tan 𝛼 tan 𝛽 = 1−𝑥𝑦
x
= tan−1 (tan x/2) = 2
𝟒 𝟏𝟐 𝟑𝟑
4) Prove that 𝐜𝐨𝐬 −𝟏 (𝟓) + 𝒄𝒐𝒔−𝟏 (𝟏𝟑) = 𝒄𝒐𝒔−𝟏 (𝟔𝟓)
𝟏
3) 𝐜𝐨𝐭 −𝟏 ( ) 𝒘𝒉𝒆𝒓𝒆 𝒙 > 𝟏 𝐶
√𝒙𝟐 −𝟏
−1 4
Solution: Let x = secθ ⇒ θ = sec x Solution: Let cos−1 ( ) = 𝐴
1 1
5 5
Now, cot −1 ( ) = cot −1 ( ) ⟹ cos 𝐴 = 5
4 3
√𝑥 2 −1 √sec2 𝜃−1
1 1 3
𝐴
= cot −1 ( )= cot −1 (𝑡𝑎𝑛𝜃) ⟹ sin 𝐴 = 5 𝐴 4 𝐵
√tan2 𝜃
= cot −1(𝑐𝑜𝑡𝜃) = 𝜃 = sec −1
𝑥
𝑍
12
Let cos−1 (13) = 𝐵
√𝟏+𝒙𝟐 −𝟏 13
4) 𝐭𝐚𝐧−𝟏 ( 𝒙
),𝒙 ≠ 𝟎 ⟹ cos 𝐵 = 13
12
5
Solution: Let 𝑥 = tanθ ⇒ θ = tan −1
x ⟹ sin 𝐵 =
5 𝐵
13 𝑋 12 𝑌
−1 √1+x2 −1 √1+tan2 θ−1
Now, tan ( x
)= tan−1 ( x
)

Mr. Sharath Patil G H, Lecturer, Dept of Mathematics, DAVANGERE Page 4

Question Bank with Solutions Inverse Trigonometric Functions

Now, Now,
cos(𝐴 + 𝐵) = cos 𝐴 cos 𝐵 − sin 𝐴 sin 𝐵 tan 𝐴+tan 𝐵
tan(𝐴 + 𝐵) = 1−tan 𝐴 tan 𝐵
4 12 3 5
⟹ cos(A + B) = . − . 8 3
+
32+45
5 13 5 13 15 4 60
48 15 ⟹ tan(A + B) = 8 3 = 60−24
⟹ cos(A + B) = 65 − 65 1− .
15 4 60
77
48−15 ⟹ tan(A + B) = 36
⟹ A + B = cos−1 [ 65 ]
77
4 12 33 ⟹ A + B = tan−1 [36]
⟹ cos−1 (5) + cos−1 (13) = cos−1 (65)
8 3 77
⟹ sin−1 ( ) + sin−1 ( ) = tan−1
17 5 36
𝟏𝟐 𝟑 𝟓𝟔
5) Prove that 𝐜𝐨𝐬 −𝟏 + 𝐬𝐢𝐧−𝟏 𝟓 = 𝐬𝐢𝐧−𝟏 𝟔𝟓
𝟏𝟑
7) Find the value of 𝒙 of the following
𝐶
12
Solution: Let cos−1 (13) = 𝐴 𝟏−𝐱 𝟏
13 1) 𝐭𝐚𝐧−𝟏 (𝟏+𝐱) = 𝟐 𝐭𝐚𝐧−𝟏 𝐱 , 𝐱 > 𝟎
12
⟹ cos 𝐴 = 5 1−𝑥 1
13 Solution : Given: tan−1 (1+𝑥) = 2 tan−1 𝑥
5 𝐴
⟹ sin 𝐴 = 13 𝐴 12 𝐵 Let, 𝑥 = tan 𝜃 ⟹ 𝜃 = tan−1 𝑥
1−tan 𝜃 1
𝑍 ⟹ tan−1 ( ) = tan−1 tan 𝜃
3 1+tan 𝜃 2
Let sin−1 ( ) =𝐵 −1 𝜋 1
5
5 ⟹ tan (tan ( 4 − 𝜃)) = 2 𝜃
3
⟹ sin 𝐵 = 3 𝜋 1 𝜋 1
5 ⟹ 4 − 𝜃 = 2 𝜃 ⟹ 4 = 2 𝜃+ 𝜃
4 𝐵
⟹ cos 𝐵 =5 𝜋 𝜃+2𝜃 𝜋 3𝜃
𝑋 4 𝑌 ⟹4= 2 ⟹4 = 2
2𝜋
⟹ 2𝜋 = 12𝜃 ⟹ 𝜃 =
Now, 12
𝜋
sin(𝐴 + 𝐵) = sin 𝐴 cos 𝐵 + cos 𝐴 sin 𝐵 ⟹ 2𝜋 = 12𝜃 ⟹ 𝜃 = 6
5 4 12 3 𝜋 1
⟹ sin(A + B) = 13 . 5 + 13
.5 Now, 𝑥 = tan 𝜃 ⟹ 𝑥 = tan 6 ⇒ 𝑥 =
√3
20 36
⟹ sin(A + B) = 65 + 65
𝝅
⟹ A + B = sin−1 [
20+36
] 2) 𝐬𝐢𝐧−𝟏 (𝟏 − 𝒙) − 𝟐 𝐬𝐢𝐧−𝟏 𝒙 = 𝟐
65 π
12 3 56 Solution: sin−1(1 − x) − 2 sin −1
x=
⟹ cos −1 + sin−1 5 = sin−1 65 2
13 π
⟹ sin−1(1 − x) = 2 + 2 sin−1 x
π
𝟖
6) Prove that 𝐬𝐢𝐧−𝟏 (𝟏𝟕) + 𝐬𝐢𝐧−𝟏 (𝟓) = 𝐭𝐚𝐧−𝟏 𝟑𝟔
𝟑 𝟕𝟕 ⟹ 1 − x = sin ( + 2 sin−1 x)
2
⟹ 1 − x = cos(2 sin−1 x)
𝐶
8 ⟹ 1 − x = 1 − 2 sin2(sin−1 x)
Solution: Let sin−1 (17) = 𝐴
17 ⟹ −x = −2[sin(sin−1 x)]2
8
⟹ sin 𝐴 = 17 8 ⟹ x = 2x 2 ⟹ x − 2x 2 = 0
8 𝐴 ⟹ x(1 − 2x) = 0
⟹ tan 𝐴 = 𝐴 15 𝐵
15
⟹ x = 0 or 1 − 2x = 0
𝑍 ⟹ x = 0 or 2x = 1 ⟹ x = 0 or x = 2
1
3
Let sin−1 (5) =𝐵 1
3
5 Since x = 2 is not a solution of the given
⟹ sin 𝐵 = 5 3
𝐵 equation, ⟹ x = 0 is the only solution.
3
⟹ tan 𝐵 = 4 𝑋 4 𝑌

Mr. Sharath Patil G H, Lecturer, Dept of Mathematics, DAVANGERE Page 5