NCERT Solutions for Class 12 Maths Chapter 2 Inverse Trigonometric Functions

NCERT
SOLUTIONS
CLASS - 12th

aglase .co

Class : 12th
Subject : Maths
Chapter : 2
Chapter Name : Inverse Trigonometric Functions

Q1 Find the principal values of the following:

( )
sin − 1 − 2
1

Answer. Let sin − 1 − 2 ( ) 1
= y.
1
Then siny = − 2 = − sin 6 () ( )
π π
= sin − 6

We know that the range of the principal value branch of sin − 1 is

[ π π
− 2, 2
] and sin − 6
( ) π
= − 2
1

Therefore, the principal value of sin − 1 − 2
( ) 1 π
is − 6 .

Page : 41 , Block Name : Exercise 2.1

Q2 Find the principal values of the following:

cos −1
()
√3
2

Answer. Let cos −1
()
√3
2
√3
= y. Then, cosy = 2 = cos 6 () π

We know that the range of the principal value branch of cos − 1 is

[0, π] and cos 6 ()π
= 2
√3

Therefore, the principal value of

cos −1
()
√3
2
π
is 6

Page : 41 , Block Name : Exercise 2.1

Q3 Find the principal values of the following:
cosec − 1(2)

Answer. Let cosec − 1(2) = y. Then cosecy = 2 = cosec 6 ()
π

We know that the range of the principal value branch of cosec − 1 is [ π π
− 2, 2 ] − {0}
.
π
Therefore, the principal value of (2) is 6 .

Page : 41 , Block Name : Exercise 2.1

Q4 Find the principal values of the following:
tan − 1( − √3)

Answer. Lettan − 1( − √3) = y. Then, tany = − √3 = − tan 3 = tan − 3
π
( ) π

We know that the range of the principal value branch of tan − 1 is

( π π
− 2, 2 ) ( )
and tan − 3
π
is − √3
π
.

Therefore, the principal value of cosec − 1(2) is 6

Page : 41 , Block Name : Exercise 2.1

Q5 Find the principal values of the following:

( )
cos − 1 − 2
1

Answer. Let cos − 1 − 2
( ) 1 1
= y. Then, cosy = − 2 = − cos 3
() ( ) ( )
π
= cos π − 3
π
= cos 3
2π

We know that the range of the principal value branch of cos − 1 is

[0, π] and cos 3
()
2π
= − 2
1

Therefore, the principal value of cos − 1 − 2 ( )
1
is 3
2π

Page : 41 , Block Name : Exercise 2.1

Q6 Find the principal values of the following:

tan − 1( − 1)

Answer. Lettan − 1( − 1) = y. Then,tany = − 1 = − tan 4 () ( )
π π
= tan − 4

We know that the range of the principal value branch of tan − 1 is

( π π
− 2, 2 ) ( )
and tan − 4
π
= −1

Therefore, the principal value of tany = − 1 = − tan 4
() ( )
π π
= tan − 4

Page : 41 , Block Name : Exercise 2.1

Q7 Find the principal values of the following:

sec − 1
()
2

√3

Answer. Let

sec − 1
()
2

√3
= y. Then , secy =
2

√3 ()
= sec 6
π

We know that the range of the principal value branch of sec − 1 is

[0, π] − {} π
2 ()
and sec 6
π
=
2

√3

Therefore, the principal value of sec − 1
() √3
2
is 6
π

Page : 42 , Block Name : Exercise 2.1

Q8 Find the principal values of the following:
cot − 1(√3)

Answer. Let

cot − 1(√3) = y. Then , coty = √3 = cot 6 . ()
π

We know that the range of the principal value branch of cot − 1 is (0,π) and cot 6
π
() √
π
= 3

Therefore, the principal value of cot − 1(√3) is 6

Page : 42 , Block Name : Exercise 2.1

Q9 Find the principal values of the following:

( )
cos − 1 −
√2
1

Answer. Let

( )
cos − 1 −
√2
1
= y. Then , cosy = −
1

√2
= − cos 4
() ( ) ( )
π π
= cos π − 4
3π
= cos 4

We know that the range of the principal value branch of cot − 1 is (0,π) and cos 4 ()
3π
= −
1

√2

Therefore, the principal value of cos − 1 −
( )
1

√2
3π
is 4 .

Page : 42 , Block Name : Exercise 2.1

Q10 Find the principal values of the following:
cosec − 1( − √2)

Answer. Let

cosec − 1( − √2) = y. Then, cosec y = − √2 = − csc 4 () ( )
π
= csc − 4
π

we know that the range of the principal value branch of cosec − 1 is

[ π π
− 2, 2 ] − {0} and cosec ( )
π
− 4 = − √2
π
Therefore, the principal value of csc − 1( − √2) is − 4

Page : 42 , Block Name : Exercise 2.1

Q11 Find the values of the following:

tan − 1(1) + cos − 1 − 2
( )
1
( )
+ sin − 1 − 2
1

Answer.
π
Let tan − 1(1) = x. Then, tanx = 1 = tan 4
π
∴ tan − 1(1) = 4

( )
Let cos − 1 − 2
1
= y. Then, cosy = − 2 = − cos 3
1
() ( ) ( )
π
= cos π − 3
π 2π
= cos 3

∴ cos − 1 − 2( ) 1
= 3
2π

( )
Let sin − 1 − 2
1 1
= z. Then, sinz = − 2 = − sin 6
() ( )
π π
= sin − 6

∴ sin − 1 − 2
( ) 1
= − 6
π

∴ tan − 1(1) + cos − 1 − 2
π 2π π
( ) 1
( )
+ sin − 1 − 2
1

= 4 + 3 − 6
3π + 8π − 2π 9π 3π
= 12
= 12 = 4

Page : 42 , Block Name : Exercise 2.1

Q12 Find the values of the following:
1 1
cos − 1 2 + 2sin − 1 2

Answer.

Let cos − 1 2 () 1 1
= x. Then, cosx = 2 = cos 3 () π

∴ cos − 1 2
() 1
= 3
π

Let sin − 1 2() 1
= y. Then, siny = 2 = sin 6
1
() π

∴ sin − 1 2 () 1
= 6
π

∴ cos − 1 2 () 1
+ 2sin − 1 2 ()
1 π 2π
= 3 + 6 = 3 + 3 = 3
π π 2π

Page : 42 , Block Name : Exercise 2.1

Q13 Find the value of if sin − 1x = y, then,
(A) 0 ≤ y ≤ π

π π
(b) − 2 ≤ y ≤ 2
(C) 0 < Y < π
π π
(D) − 2 < y < 2

Answer. It is given that sin − 1x = y

We know that the range of the principal value branch of sin − 1 is [ π π
− 2, 2 ]
π π
Therefore, − 2 ≤ y ≤ 2

Page : 42 , Block Name : Exercise 2.1

Q14 Find the value of tan − 1√3 − sec − 1( − 2) is equal to
(A) π
π
(B) − 3
π
(C) 3
2π
(D) 3

π
Answer. Let tan − 1√3 = x. Then, tan x = √3 = tan 3

We know that the range of the principal value branch of tan − 1 is
( ) −π π
2
,2

π
∴ tan − 1√3 = 3

Let sec − 1( − 2) = y. Then, sec y = − 2 = − sec 3 () ( )
π π
= sec π − 3
2π
= sec 3

We know that the range of the principal value branch of sec − 1 is [0, π] − {} π
2

2π
∴ sec − 1( − 2) = 3
π 2π π
Hence tan − 1(√3) − sec − 1( − 2) = 3 − 3 = − 3

Page : 42 , Block Name : Exercise 2.1

Q1 Prove the following:

(
3sin − 1x = sin − 1 3x − 4x 3 , x ∈ ) [ 1 1
− 2, 2 ]

Answer. To prove

( 3x − 4x ) , x ∈ [ − , ]
1 1
3sin − 1 x = sin − 1 3
2 2

Let x = sinθ. Then , sin − 1x = θ
We have,

( )
R.H.S. = sin − 1 3x − 4x 3 = sin − 1 3sinθ − 4sin 3θ ( )
= sin − 1(sin3θ)
= 3θ
= 3sin − 1x
= L. H. S

Page : 47 , Block Name : Exercise 2.2

Q2 Prove the following:

(
3cos − 1x = cos − 1 4x 3 − 3x , x ∈ ) [ ]
1
2
,1

Answer. To prove

(
3cos − 1x = cos − 1 4x 3 − 3x , x ∈ ) [ ]
1
2
,1

Let x = cosθ. Then , cos − 1x = θ
we have.

R.H.S. = cos − 1 4x 3 − 3x( )
(
= cos − 1 4cos 3θ − 3cosθ )
= cos − 1(cos3θ)
= 3θ
= 3cos − 1x
= L.H.S.

Page : 47 , Block Name : Exercise 2.2

Q3 Prove the following:
2 7 1
tan − 1 11 + tan − 1 24 = tan − 1 2

Answer. To prove
2 1 1
tan − 1 11 + tan − 1 24 = tan − 1 2

2 7
[
= tan − 1 11 + tan − 1 24 tan − 1x + tan − 1y = tan − 1 1 − xy
x+y
]

2 7
11 + 24
= tan − 1 2 7
1 − 11 ⋅ 24
48 + 77
11 × 24
= tan − 1 11 × 24
11 × 24
48 + 77 125 1
= tan − 1 264 − 14 = tan − 1 250 = tan − 1 2 = R. H. S.

Page : 47 , Block Name : Exercise 2.2

Q4 Prove the following:
1 1 31
2tan − 1 2 + tan − 1 7 = tan − 1 17

Answer. To prove
1 1 31
2tan − 1 2 + tan − 1 7 = tan − 1 17
1 1
L.H.S. = 2tan − 1 2 + tan − 1 7

[ ]
1
2⋅ 2 1 2x
= tan − 1 + tan − 1 7 2tan − 1x = tan − 1
1 − x2
1−
() 1
2
2

1 1
= tan − 1 + tan − 1 7
()
4
3
4
1

[ ]
3
+7 x+y
= tan − 1 4 1 tan − 1x + tan − 1y = tan − 1 1 − xy
1− 3 ⋅ 7

= tan
( )−1
28 + 3
21

( ) 21 − 4
21

31
= tan − 1 17 = R. H. S

Page : 47 , Block Name : Exercise 2.2

Q5 Write the following functions in the simplest form
√1 + x 2 − 1
tan − 1 x
,x ≠ 0

Answer.

√ 1 + x2 − 1
−1
tan x

Put x = tanθ ⇒ θ = tan − 1x

∴ tan −1
√1 + x 2 − 1
x = tan −1
( √1 + tan2 θ − 1
tan θ )
= tan − 1 ( sec θ − 1
tan θ ) = tan − 1 ( 1 − cos θ
sin θ )

( )
θ
2sin 2 2
= tan − 1 θ θ
2sin 2 cos 2

( )
= tan − 1 tan 2
θ θ 1
= 2 = 2 tan − 1x

Page : 47 , Block Name : Exercise 2.2

Q6 Write the following functions in the simplest form
1
tan − 1 , |x| > 1
√x 2 − 1

Answer.
1
tan − 1 , |x| > 1
√x 2 − 1
Put x = cosecθ ⇒ θ = cosec − 1x
1 1
∴ tan − 1 = tan − 1
√x 2 − 1 √cos ec2θ − 1 [ π
cosec − 1x + sec − 1x = 2

( )
= tan − 1 cot θ
1
= tan − 1(tanθ)

π
= θ = cosec − 1x = 2 − sec − 1x

Page : 47 , Block Name : Exercise 2.2

Q7 Write the following functions in the simplest form

tan − 1
(√ ) 1 − cos x
1 + cos x
,0 < x < π

Answer.

tan − 1
(√ ) 1 − cos x
1 + cos x
,x < π

tan − 1
(√ ) 1 − cos x
1 + cos x
= tan − 1
(√ )2cos 2 2
x
2sin 2 2
x

= tan − 1

x
() () x
sin 2

cos 2
x = tan − 1 tan 2
x

= 2

Page : 47 , Block Name : Exercise 2.2

Q8 Write the following functions in the simplest form

( cos x − sin x
)
tan − 1 cos x + sin x , 4 < x < 4
−π 3π

Answer.

( cos x − sin x
tan − 1 cos x + sin x )

( )
sin x
1 − cos x
= tan − 1 sin x
1 + cos x

(
= tan − 1 1 + tan x
1 − tan x
)
= tan − 1(1) − tan − 1(tanx)
π
[ x−y
tan − 1 1 − xy = tan − 1x − tan − 1y ]
= 4 −x

Page : 47 , Block Name : Exercise 2.2

Q9 Write the following functions in the simplest form
x
tan − 1 , |x| < a
√a 2 − x 2

Answer.
x
tan − 1
√a 2 − x 2
x
Put x = asinθ ⇒ a = sinθ ⇒ θ = sin − 1 a () x

∴ tan − 1
x

√a 2 − x 2
= tan − 1
( asin θ

√a2 − a2sin2 θ )
= tan − 1
( asin θ

a 1 − sin 2 θ
√ ) = tan − 1 acos θ ( )
asin θ

x
= tan − 1(tanθ) = θ = sin − 1 a

Page : 47 , Block Name : Exercise 2.2

Q10 Write the following functions in the simplest form

tan −1
( )
3a 2x − x 3
a 3 − 3ax 2
, a > 0;
−a

√3
<x<
√3
a

Answer.

tan − 1
( )
3a 2x − x 3
a 3 − 3ax 2

x x
Put x = atanθ ⇒ a = tanθ ⇒ θ = tan − 1 a

tan −1
( ) (
3a 2x − x 3
a 3 − 3ax 2
= tan −1
3a 2 ⋅ atan θ − a 3tan 3 θ
a 3 − 3a ⋅ a 2tan 2 θ )
= tan
( )
−1
3tan θ − tan 3 θ
1 − 3tan 2 θ

= tan − 1
( ) 3tan θ − tan 3 θ
1 − 3tan 2 θ

= tan − 1(tan3θ)
= 3θ
x
= 3tan − 1 a

Page : 47 , Block Name : Exercise 2.2

Q11 Find the values of each of the following:

[ (
tan − 1 2cos 2sin − 1 2
1
)]
1
Answer. Let sin − 1 2 = x. Then,sinx = 2 = sin 6
1 π
1
()π

∴ sin − 1 2 = 6

[ (
∴ tan − 1 2cos 2sin − 1 2
1
)] [ ( )]
= tan − 1 2cos 2 × 6
π

= tan − 1 2cos 3 [ ] π
= tan − 1 2 × 2[ ] 1

π
= tan − 11 = 4

Page : 47 , Block Name : Exercise 2.2

Q12 Find the values of each of the following:

(
cot tan − 1a + cot − 1a )
(
Answer. cot tan − 1a + cot − 1a )
= cot 2
() [
π
tan − 1x + cot − 1x = 2
π
]
=0

Page : 47 , Block Name : Exercise 2.2

Q13 Find the values of each of the following:

tan 2
1
[ sin − 1
2x
1 + x2
+ cos − 1
1 − y2
1 + y2 ] , | x | < 1, y > 0 and xy < 1

Answer. Let x = tanθ. Then , θ = tan − 1x

∴ sin − 1
2x
1 + x2
= sin − 1
( ) 2tan θ
1 + tan 2 θ
= sin − 1(sin2θ) = 2θ = 2tan − 1x

Let y = tanΦ. Then , Φ = tan − 1y

⋅ cos − 1
1 − y2
1+y 2 = cos
−1
( ) 1 − tan 2 ϕ
1 + tan 2 ϕ
= cos − 1(cos2ϕ) = 2ϕ = 2tan − 1y

1
∴ tan 2 sin
[ −1
2x
1 + x2
+ cos −1
1 − y2
1 + y2 ]
[ ]
1
= tan 2 2tan − 1x + 2tan − 1y

[
= tan tan − 1x + tan − 1y ]
[ ( )]
= tan tan − 1 1 − xy
x+y
x+y

= 1 − xy

Page : 47 , Block Name : Exercise 2.2

Q14 Find the values of each of the following:

( 1
)
sin sin − 1 5 + cos − 1x = 1, then find the value of x.

Answer.

( 1
sin sin − 1 5 + cos − 1x = 1 )
( 1
) ( )
⇒ sin sin − 1 5 cos cos − 1x + cos sin − 1 5 sin cos − 1x = 1 ( 1
) ( )
[sin(A + B) = sinAcosB + cosAsinB]
1
(
⇒ 5 × x + cos sin − 1 5 sin cos − 1x = 1
1
) ( )
x
( 1
) (
⇒ 5 + cos sin − 1 5 sin cos − 1x = 1 .......(1)
1
)
Now, let sin − 1 5 = y

√ () ( )
1 1 2 2√ 6 2√ 6
Then,siny = 5 ⇒ cosy = 1− 5
= 5 ⇒ y = cos − 1 5

( )
1 2√ 6
∴ sin − 1 5
= cos − 1 5
......(2)

Let cos − 1x = z

Then, cosz = x ⇒ sinz = √1 − x 2 ⇒ z = sin − 1 (√1 − x 2 )
∴ cos − 1x = sin − 1 (√ )
1 − x 2 .....(3)

From (1),(2), and (3) we have:

( ) ( √
2√ 6
x
5
+ cos cos − 1 5
⋅ sin sin − 1 1 − x 2 = 1 )
x 2√ 6
⇒ 5 + 5 ⋅ √1 − x 2 = 1
√
⇒ x + 2√6 1 − x 2 = 5

√
⇒ 2√6 1 − x 2 = 5 − x
On squaring both sides,we get

( )
(4)(6) 1 − x 2 = 25 + x 2 − 10x

⇒ 24 − 24x 2 = 25 + x 2 − 10x
⇒ 25x 2 − 10x + 1 = 0
⇒ (5x − 1) 2 = 0
⇒ (5x − 1) = 0
1
⇒x= 5
1
Hence, the value of x is 5

Page : 47 , Block Name : Exercise 2.2

Q15 Find the values of each of the following:
x−1 x+1 π
tan − 1 x − 2 + tan − 1 x + 2 = 4 , then find the value of x

x−1 x+1 π
Answer. tan − 1 x − 2 + tan − 1 x + 2 = 4

[ ] [
x−1 x+1

]
x−2
+ x+2 π x+y
⇒ tan − 1 = 4 tan − 1x + tan − 1y = tan − 1 1 − xy
1− ( )( )
x−1
x−2
x+1
x+2

[ (x−1) (x+2) + (x+1) (x−2)
⇒ tan − 1 ( x + 2 ) ( x − 2 ) − ( x − 1 ) ( x + 1 )
] π
= 4

⇒ tan − 1
[ ] x2 + x − 2 + x2 − x − 2
x2 − 4 − x2 + 1
= 4
π

⇒ tan − 1
[ ] 2x 2 − 4
−3
= 4
π

[ ]
⇒ tan tan − 1
4 − 2x 2
3
π
= tan 4

4 − 2x 2
⇒ 3
=1

⇒ 4 − 2x 2 = 3
⇒ 2x 2 = 4 − 3 = 1
1
⇒x= ±
√2
1
Hence the value of x is. ±
√2

Page : 47 , Block Name : Exercise 2.2

Q16 Find the values of each of the expressions sin − 1 sin 3 . ( ) 2π

Answer. sin − 1 sin 3 ( ) 2π

We know that sin − 1(sinx) = x if x ∈
[ π π
]
− 2 , 2 , which is the principal value branch of

sin − 1x.

Here, 3 ∉
2π
[ ] −π π
2 , 2

( )
Now, sin − 1 sin 3
2π
can be written as,

( ) [ ( )]
sin − 1 sin 3
2π
= sin − 1 sin π − 3
2π
( )
= sin − 1 sin 3
π π
where 3 ∈ [ ]
−π π
2 , 2

( ) ( )
∴ sin − 1 sin 3
2π
= sin − 1 sin 3
π π
= 3

Page : 47 , Block Name : Exercise 2.2

Q17 Find the values of each of the expressions

( )
tan − 1 tan 4
3π

Answer. tan − 1 tan 4 ( ) 3π

We know that tan − 1(tanx) = x if x ∈ ( π π
)
− 2 , 2 , which is the principal value branch of

tan − 1x.

3π
Here, 4 ∉
( ) −π π
2
,2

( )
Now, tan − 1 tan 4
3π

can be written as:

( ) [ ( )] [ ( )]
tan − 1 tan 4
3π
= tan − 1 − tan
− 3π
4
= tan − 1 − tan π − 4
π

[ ] [ ( )]
= tan − 1 − tan 4
( )π
= tan − 1 tan − 4
π
where − 4 ϵ
π −π π
2
,2

( ) [ ( )]
∴ tan − 1 tan 4
3π
= tan − 1 tan
−π
4
= 4
−π

Page : 47 , Block Name : Exercise 2.2

Q18 Find the values of each of the expressions

( 3
tan sin − 1 5 + cot − 1 2
3
)
3 3 4 5
Answer. Let sin − 1 5 = x . Then x = 5 ⇒ cosx = √ 1 − sin 2x = 5 ⇒ secx = 4
25 3
∴ tanx = √ sec 2x − 1 =
√ 16 − 1 = 4

3
∴ x = tan − 1 4

3 3
∴ sin − 1 5 = tan − 1 4 … (i)

3
Now, cot − 1 2 = tan − 1 3
2
… (ii) [ 1
tan − 1 x = cot − 1x ]
(
Hence, tan sin − 1 5 + cot − 1 2
3 3
)
( 3
= tan tan − 1 4 + tan − 1 3
2
)
(
= tan tan − 1 4 + 3
3 2
)
(
= tan tan − 1 12 − 6
9+8
)
(
= tan tan − 1 6
17
) = 6
17

(
= tan tan − 1 6
17
) = 6
17

Page : 47 , Block Name : Exercise 2.2

Q19 Find the value of cos − 1 cos 6
7π
( ) 7π
is equal to.

(A) 6
5π
(B) 6
5π
(C) 6
π
(D) 6

Answer. We know that cos − 1(cosx) = x if x ∈ [0, π], ,which is the principal value branch of cos − 1(x).

7π
Here 6 ∉ x ∈ [0, π]

Now,cos − 1 cos 6 ( ) 7π
can be written as,

( ) ( )
cos − 1 cos 6
7π
= cos − 1 cos 6
− 7π
[ (
= cos − 1 cos 2π − 6
7π
)] [cos(2π + x) = cosx]

[ ]
= cos − 1 cos 6
5π 5π
where 6 ϵ[0, π]

( ) ( )
∴ cos − 1 cos 6
7π
= cos − 1 cos 6
5π
= 6
5π

The correct answer is B.

Page : 47 , Block Name : Exercise 2.2

( π
Q20 sin 3 − sin − 1 − 2
1
( )) 1
is equal to

(A) 2
1
(B) 3
1
(C) 4
(D) 1

Answer. Let sin − 1
( )
−1
2
=x
−1
. Then x = 2 = − sin 6 = sin
π
( )
−π
6

We know that the range of the principal value branch of sin − 1 is
[ ]
−π π
2
,2

sin − 1 ( )
−1
2
= 6
−π

( ( )) ( ) ( ) ( )
π
∴ sin 3 − sin − 1
−1
2
= sin 3 + 6
π π
= sin 6
3π
= sin 2
π
=1

The correct answer is D.

Page : 47 , Block Name : Exercise 2.2

Q21 tan − 1√3 − cot − 1( − √3) is equal to:
(A) π
π
(B) − 2
(C) 0
(D) 2√3

Answer. Let tan − 1√3 = x. Then,

π
tanx = √3 = tan 3 where 3 ∈
π
( π π
− 2, 2 )
We know that the range of the principal value branch of

tan − 1 is
( π π
− 2, 2
)
π
∴ tan − 1√3 = 3

Let − 1( − √3) = y

coty = − √3 = − cot 6
() ( )
π
= cot π − 6
π 5π
= cot 6 where 6 ∈ (0, π)
5π

The range of the principal value branch of cot − 1 is (0, π)
5π
∴ cot − 1( − √3) = 6
π 5π 2π − 5π − 3π π
∴ tan − 1√3 − cot − 1( − √3) = 3 − 6 = 6
= 6
= − 2
The correct answer is B.

Page : 47 , Block Name : Exercise 2.2

Q1 Find the value of the following:

( )
cos − 1 cos 6
13π

Answer. We know that cos − 1(cosx) = x if x ∈ [0, π] which is the principal value branch of,cos − 1x
13π
Here 6 ∉ [0, π]

( )
Now, cos − 1 cos 6
13π
can be written as:

( ) [ ( )]
cos − 1 cos 6
13π
= cos − 1 cos 2π + 6
π
= cos − 1 cos 6 [ ( )] π π
, where 6 ∈ [0, π]

( ) [ ( )]
∴ cos − 1 cos 6
13π
= cos − 1 cos 6
π
= 6
π

Page : 51 , Block Name : Miscellaneous Exercise

Q2 Find the value of the following:tan − 1 tan 6 ( ) 7π

Answer.

We know that tan − 1(tanx) = x if x ∈
( π π
)
− 2 , 2 , which is the principal value branch of

tan − 1x.
7π
Here, 6 ∉
( ) π π
− 2, 2

( )
Now, tan − 1 tan 6
7π
can be written as,

( ) [ ( )]
7π
tan − 1 tan 6 = tan − 1 tan 2π − 6
5π
= tan(2π − x) = − tanx

[ ( )] [ ( )]
= tan − 1 − tan 6
5π
= tan − 1 tan − 6
5π
[ ) 5π
= tan − 1 π − 6

[ ( )]
= tan − 1 tan 6( ) π
, where 6 ∈
π π π
− 2, 2

( ) ( )
∴ tan − 1 tan 6
7π
= tan − 1 tan 6
π
= 6
π

Page : 51 , Block Name : Miscellaneous Exercise

3 24
Q3 Prove that2sin − 1 5 = tan − 1 7

Answer.
3 3
Let sin − 1 5 = x. Then, sinx = 5

⇒ cosx =
√ () 1−
3
5
2
= 5
4

3
∴ tanx = 4
3 3 3
∴ x = tan − 1 4 ⇒ sin − 1 5 = tan − 1 4

3 3
L.H.S. = 2sin − 1 = 2tan − 1
5 4

( ())
3
2× 4
= tan − 1
3 2
1− 4

() ( )
3
2 3 16
= tan − 1 16 − 9 = tan − 1 ×
2 7
16

24
= tan − 1 7 = R. H. S

Page : 51 , Block Name : Miscellaneous Exercise

Q4 Prove that
8 3 77
sin − 1 17 + sin − 1 5 = tan − 1 36

Answer.

8
Let sin − 1 17 = x. Then, sinx = 17 ⇒ cosx =
8
√ () √
1−
8
17
2
=
225 15
289 = 17

8 8
∴ tanx = 15 ⇒ x = tan − 1 15
8 8
∴ sin − 1 17 = tan − 1 15 . . . . . . (1)

3
Now , let sin − 1 5 = y. Then,siny = 5 ⇒ cosy =
3
√ () √
1−
3
5
2
=
16
25
4
= 5.

3 3
∴ tany = 4 ⇒ y = tan − 1 4
3 3
∴ sin − 1 5 = tan − 1 4 . . . . . . (2)
Now we have :
8 3
L.H.S = sin − 1 17 + sin − 1 5
8 3
= tan − 1 15 + tan − 1 4
Using (1) and (2)
8 3
15
+4
= tan − 1 8 3
1 − 15 × 4

= tan − 1 ( 32 + 45
60 − 24 )
77
= tan − 1 = R. H. S.
36

Page : 51 , Block Name : Miscellaneous Exercise

4 12 33
Q5 Prove that cos − 1 5 + cos − 1 13 = cos − 1 65

Answer.
4
Let cos − 1 5 = x. Then, cosx = 5 ⇒ sinx =
4
√ () 1−
4
5
2 3
= 5

3 3
∴ tanx = 4 ⇒ x = tan − 1 4
4 3
∴ cos − 1 5 = tan − 1 4 . . . . . (1)
12 12 5
Now, let cos − 1 13 = y. Then, cos y = 13 ⇒ siny = 13
5 5
∴ tany = 12 ⇒ y = tan − 1 12
12 5
∴ cos − 1 13 = tan − 1 12 . . . . (2)
33 33 56
Let cos − 1 65 = z. Then, cosz = 65 ⇒ sinz = 65
56 56
∴ tanz = 33 ⇒ z = tan − 1 33
33 56
∴ cos − 1 65 = tan − 1 33 . . . . . (3)
Now we will prove that:
4 12
L. H.S. = cos − 1 + cos − 1
5 13
[ Using (1) and (2)]
3 5
= tan − 1 + tan − 1
4 12
3 5

= tan − 1
4 + 12
3
1 − 4 ⋅ 12
5 [ x+y
tan − 1x + tan − 1y = tan − 1 1 − xy ]
36 + 20
= tan − 1 48 − 15
56
= tan − 1 33
56
= tan − 1 33 [by (3)]
=R.H.S

Page : 51 , Block Name : Miscellaneous Exercise

Q6 Prove that
12 3 56
cos − 1 13 + sin − 1 5 = sin − 1 65

Answer.
3
Let sin − 1 5 = x. Then, sinx = 5 ⇒ cosx =
3
√ () √
1−
3
5
2
=
16
25
4
= 5

3 3
∴ tanx = 4 ⇒ x = tan − 1 4
3 3
∴ sin − 1 5 = tan − 1 4 . . . . (1)
12 12 5
Now, let cos − 1 13 = y. Then, cosy = 13 ⇒ siny = 13
5 5
∴ tany = 12 ⇒ y = tan − 1 12
12 5
∴ cos − 1 13 = tan − 1 12 . . . . . (2)
56 56 33
Let sin − 1 65 = z. Then, sinz = 65 ⇒ cosz = 65
56 56
∴ tanz = 33 ⇒ z = tan − 1 33
56 56
∴ sin − 1 65 = tan − 1 33 . . . . (3)
12 3
L.H.S. = cos − 1 + sin − 1
13 5
[ Using (1) and (2)]
−1
5 −1
3
= tan + tan
12 4
5 3

= tan − 1
12 + 4
5
1 − 12 ⋅ 4
3 [ tan − 1x + tan − 1y = tan − 1 1 − xy
x+y
]
20 + 36
= tan − 1
48 − 15
−1
56
= tan
33
56
= sin − 1 = R. H. S
65

Page : 51 , Block Name : Miscellaneous Exercise

63 5 3
Q7 Prove that tan − 1 16 = sin − 1 13 + cos − 1 5 .

Answer.

5 5 12
Let sin − 1 13 = x. Then, sinx = 13 ⇒ cosx = 13
5 5
∴ tanx = 12 ⇒ x = tan − 1 12
5 5
∴ sin − 1 13 = tan − 1 12 . . . . . . (1)
Using (1) and (2), we have
5 3
R.H.S. = sin − 1 13 + cos − 1 5
5 4
= tan − 1 12 + tan − 1 3 . . . . (2)

( )[
5 4

]
12
+3 x+y
= tan − 1 5 4 tan − 1x + tan − 1y = tan − 1 1 − xy
1 − 12 × 3

= tan − 1
( 15 + 48
36 − 20 )
63
= tan − 1
16
= L. H. S

Page : 51 , Block Name : Miscellaneous Exercise

Q8 Prove that
1 1 1 1 π
tan − 1 5 + tan − 1 7 + tan − 1 3 + tan − 1 8 = 4

Answer.
1 1 1 1
= tan − 1 5 + tan − 1 7 + tan − 1 3 + tan − 1 8

= tan − 1
( ) ( )[
1
5+7

1− 5 × 7
1
1

1 + tan − 1
1
3+8

1− 3 × 8
1
1

1
x+y
tan − 1x + tan − 1y = tan − 1 1 − xy
]

= tan − 1
( 7+5
35 − 1 ) + tan − 1
( 8+3
24 − 1 )
12 11
= tan − 1 + tan − 1
34 23
−1
6 −1
11
= tan + tan
17 23

( )
6 11
17
+ 23
= tan − 1 6 11
1 − 17 × 23

( )
= tan − 1 391 − 66
138 + 187

= tan − 1 325 ( ) 325
= tan − 11

π
= 4 = R. H. S

Page : 51 , Block Name : Miscellaneous Exercise

Q9 Prove that
1 1
tan − 1√x = 2 cos − 1 1 + x , x ∈ [0, 1]

Answer.
Let x = tan 2θ. Then, √x = tanθ ⇒ θ = tan − 1√x
1−x 1 − tan 2 θ
∴ 1+x = = cos2θ
1 + tan 2 θ

Now, we have:

1
R. H.S. = 2 cos − 1 1 + x
( )
1−x 1 1
= 2 cos − 1(cos2θ) = 2 × 2θ = θ = tan − 1√x = L.H.S.

Page : 52 , Block Name : Miscellaneous Exercise

Q10 Prove that

cot − 1
( √1 + sin x
√1 + sin x − √1 − sin x ) x
( )
= 2 , x ∈ 0, 4
π

Answer.

√1 + sin x + √1 − sin x
Consider
√1 + sin x − √1 − sin x
( √1 + sin x + √1 − sin x ) 2
= (by rationalizing)
( √1 + sin x ) 2 − ( √1 − sin x ) 2

( 1 + sin x ) + ( 1 − sin x ) + 2√ ( 1 + sin x ) ( 1 − sin x )
= 1 + sin x − 1 + sin x

( √
2 1+ 1 − sin 2x ) 1 + cosx 2cos 2 2
x

= = = x x
2sinx sinx
2sin 2 cos 2
x
= cot
2

∴ L ⋅ H ⋅ S = cot − 1
( √1 + sin x + √1 − sin x
√1 + sin x − √1 − sin x ) ( )
x
= cot − 1 cot 2
x
= 2 = R. H. S.

Page : 52 , Block Name : Miscellaneous Exercise

Q11 Prove that

tan − 1
( √1 + x − √1 − x
√1 + x + √1 − x ) π 1
= 4 − 2 cos − 1x, −
√2
1
≤ x ≤ 1[ Hint: Put x = cos2θ]

Answer.
1
Put x = cos2θ so that θ = 2 cos − 1x . Then, we have:

L. H.S. = tan − 1
( √1 + x − √1 − x
√1 + x + √1 − x )
= tan − 1
( √1 + cos 2θ − √1 − cos 2θ
√1 + cos 2θ + √1 − cos 2θ )
( )
√ 2cos 2 θ − √2sin 2 θ
−1
= tan
√2cos2 θ + √2sin2 θ

= tan − 1
( √2cos θ − √2sin θ
√2cos θ + √2sin θ )
( cos θ − sin θ
= tan − 1 cos θ + sin θ ) ( 1 − tan θ
= tan − 1 1 + tan θ )

= tan − 11 − tan − 1(tanθ) tan − 1 1 + xy
π π 1
[ ( ) x−y
= tan − 1x − tan − 1y ]
= 4 − θ = 4 − 2 cos − 1x = R. H. S.

Page : 52 , Block Name : Miscellaneous Exercise

Q12 Prove that
9π 9 1 9 2√ 2
8
− 4 sin − 1 3 = 4 sin − 1 3

Answer.
9π 9 1
L.H.S. = − sin − 1
8 4 3

=
(
9 π
4 2
− sin − 1
1
3 )
9
(
= 4 cos − 1 3
1
) [
…(1) sin − 1x + cos − 1x = 2
π
]
1
Now, let cos − 1 3 = x. Then, cosx = 3 ⇒ sinx =
1
√ ()
1−
1
3
= 3
2√ 2

2√ 2 1 2√ 2
∴ x = sin − 1 3 ⇒ cos − 1 3 = sin − 1 3
9 2√ 2
∴ L. H. S. = 4 sin − 1 3 = R. H. S.

Page : 52 , Block Name : Miscellaneous Exercise

Q13 Solve the following equations:
2tan − 1(cosx) = tan − 1(2 cosecx)

Answer.
2tan − 1(cosx) = tan − 1(2 cosecx)

⇒ tan − 1
( )
2cos x
1 − cos 2 x
= tan − 1(2 cosecx)
[
2tan − 1x = tan − 1
2x
1 − x2 ]

2cos x
⇒ = 2 cosecx
1 − cos 2 x
2cos x 2
⇒ = sin x
sin 2 x
⇒ cosx = sinx
⇒ tanx = 1
π
x= 4

Page : 52 , Block Name : Miscellaneous Exercise

1−x 1
Q14 Solve the following equations:tan − 1 1 + x = 2 tan − 1x, (x > 0)

1−x 1
Answer. tan − 1 1 + x = 2 tan − 1x

⇒ tan − 11 − tan − 1x = 2 tan − 1x
π 3
1
[ x−y
tan − 1x − tan − 1y = tan − 1 1 + xy
]
⇒ 4 = 2 tan − 1x
π
⇒ tan − 1x = 6
π
⇒ x = tan 6
1
∴x=
√3

Page : 52 , Block Name : Miscellaneous Exercise

(
Q15 sin tan − 1x , | x | < 1 )
x
(A)
√1 − x 2
1
(b)
√1 − x 2
1
(C)
√1 + x 2
x
(D)
√1 + x 2
x
Answer. Let tan − 1x = y.Then tany = x ⇒ siny =
√1 + x 2

( )
∴ y = sin − 1
x

√1 + x 2
⇒ tan − 1x = sin − 1
( ) x

√1 + x 2

( ) ( )
x x
∴ sin tan − 1x = sin sin − 1 =
√1 + x 2 √1 + x 2
The correct answer is D.

Page : 52 , Block Name : Miscellaneous Exercise

π
Q16 sin 2(1 − x) − 2sin − 1x = 2 then x is equalto
i
(A) 0, 2
1
(B) 1, 2
(C) 0
1
(D) 2

Answer.
π
sin − 1(1 − x) − 2sin − 1x = 2
π
⇒ − 2sin − 1x = 2 − sin − 1(1 − x)

⇒ − 2sin − 1x = cos − 1(1 − x). . . . . (1)
Let sin − 1x = θ ⇒ sinθ = x ⇒ cosθ = √1 − x 2
∴ θ = cos − 1 (√ 1 − x2 )
∴ ∴ sin − 1x = cos − 1 (√ 1 − x2 )
Therefore, from equation (1), we have

− 2cos − 1 (√ )
1 − x 2 = cos − 1(1 − x)

Put x = siny. Then, we have:

− 2cos − 1 (√ )
1 − sin 2y = cos − 1(1 − siny)

⇒ − 2cos − 1(cosy) = cos − 1(1 − siny)

− 2cos − 1 (√ )
1 − sin 2y = cos − 1(1 − siny)

⇒ − 2cos − 1(cosy) = cos − 1(1 − siny)
⇒ − 2y = cos − 1(1 − siny)
⇒ 1 − siny = cos( − 2y) = cos2y
⇒ 1 − siny = 1 − 2sin 2y
⇒ siny(2siny − 1) = 0
⇒ siny(2siny − 1) = 0
⇒ siny(2siny − 1) = 0
1
⇒ siny = 0 or 2
1
∴ x = 0 or x = 2
1
But, when x = 2 , , it can be observed that:

L.H.S. = sin − 1 1 −
( ) 1
2
− 2sin − 1
1
2

= sin − 1 () 1
2
− 2sin − 1
1
2
1
= − sin − 1
2
π π
= − ≠ ≠ R. H. S
6 2
1
∴x= is not the solution of the given equation.
2
Thus, x = 0
Hence, the correct answer is C .

Page : 52 , Block Name : Miscellaneous Exercise

Q17 tan − 1 7
π
()
x

π
x−y
− sin − 1 x + y , is equal to:
π − 3π
(A) 2 (B) 3 (C) 4 (D) 4

Answer. tan − 1 y () x
− tan − 1 x + y
x−y

[ ][
x x−y

]
−
y x+y x−y
= tan − 1 tan − 1y − tan − 1y = tan − 1 1 + xy
1+
( )( )
x
y
x−y
x+y

[ ]
x(x+y) −y(x−y)
y(x+y)
= tan − 1 y(x+y) +x(x−y)
y(x+y)

= tan − 1
( )
x 2 + xy − xy + y 2
xy + y 2 + x 2 − xy

= tan − 1
( )
x2 + y2
x2 + y2
= tan − 11 = 4
π

Here the correct answer is C.

Page : 52 , Block Name : Miscellaneous Exercise