NCERT Solutions for Class 12 Maths Chapter 2 Inverse Trigonometric Functions Ex 2.2

NCERT Solutions for Class 12 Maths Chapter 2 Inverse Trigonometric Functions Ex 2.2 are part of NCERT Solutions for Class 12 Maths. Here we have given NCERT Solutions for Class 12 Maths Chapter 2 Inverse Trigonometric Functions Ex 2.2.

Inverse Trigonometric Functions Class 12 Ex 2.1

Board	CBSE
Textbook	NCERT
Class	Class 12
Subject	Maths
Chapter	Chapter 2
Chapter Name	Inverse Trigonometric Functions
Exercise	Ex 2.2
Number of Questions Solved	21
Category	NCERT Solutions

NCERT Solutions for Class 12 Maths Chapter 2 Inverse Trigonometric Functions Ex 2.2

Question 1.
$3\sin ^{ -1 }{ x=\sin ^{ -1 }{ (3x-4x^{ 3 });x\in \left[ -\frac { 1 }{ 2 } ,\frac { 1 }{ 2 } \right] } }$
Solution:
Let sin^-1 x = θ
sin θ = x sin 3θ = 3 sin θ – 4 sin³ θ
sin 3θ = 3x – 4x³
3θ = sin^-1(3x – 4x³)
or $3\sin ^{ -1 }{ x=\sin ^{ -1 }{ (3x-4x^{ 3 });x\in \left[ -\frac { 1 }{ 2 } ,\frac { 1 }{ 2 } \right] } }$

Question 2.
$3\cos ^{ -1 }{ x } =\cos ^{ -1 }{ \left( { 4x }^{ 3 }-3x \right) ,x\in \left[ \frac { 1 }{ 2 } ,1 \right] }$
Solution:
Let cos^-1 x = θ
x = cos θ
R.H.S= cos^-1 (4x³ – 3cosx)
= cos^-1 (4 cos³θ – 3 cosθ)
= cos^-1 (cos 3θ) [∴ cos 3θ = 4 cos³ θ – 3 cos θ]
= 3θ
= 3 cos^-1 x
= L.H.S.

Question 3.
$\tan ^{ -1 }{ \frac { 2 }{ 11 } } +\tan ^{ -1 }{ \frac { 7 }{ 24 } } =\tan ^{ -1 }{ \frac { 1 }{ 2 } }$
Solution:
L.H.S = $\tan ^{ -1 }{ \frac { 2 }{ 11 } } +\tan ^{ -1 }{ \frac { 7 }{ 24 } }$
= $\tan ^{ -1 }{ \left[ \frac { \frac { 2 }{ 11 } +\frac { 7 }{ 24 } }{ 1-\frac { 2 }{ 11 } \times \frac { 7 }{ 24 } } \right] }$
= $\tan ^{ -1 }{ \left[ \frac { 1 }{ 2 } \right] }$
= R.H.S

Question 4.
$2\tan ^{ -1 }{ \frac { 1 }{ 2 } } +\tan ^{ -1 }{ \frac { 1 }{ 7 } } =\tan ^{ -1 }{ \frac { 31 }{ 17 } }$
Solution:
L.H.S =
$2\tan ^{ -1 }{ \frac { 1 }{ 2 } } +\tan ^{ -1 }{ \frac { 1 }{ 7 } }$
NCERT Solutions for Class 12 Maths Chapter 2 Inverse Trigonometric Functions 4

Question 5.
Write the function in the simplest form
$\tan ^{ -1 }{ \left( \frac { \sqrt { 1+{ x }^{ 2 }-1 } }{ x } \right) } ,x\neq 0$
Solution:
Putting x = θ
∴ θ = tan^-1 x
NCERT Solutions for Class 12 Maths Chapter 2 Inverse Trigonometric Functions 5

Question 6.
$\tan ^{ -1 }{ \left( \frac { 1 }{ \sqrt { { x }^{ 2 }-1 } } \right) ,\left| x \right| } >1$
Solution:
Given expression
$\tan ^{ -1 }{ \left( \frac { 1 }{ \sqrt { { x }^{ 2 }-1 } } \right) ,\left| x \right| } >1$
Let x = secθ
NCERT Solutions for Class 12 Maths Chapter 2 Inverse Trigonometric Functions 6

Question 7.
$\tan ^{ -1 }{ \left( \sqrt { \frac { 1-cosx }{ 1+cosx } } \right) } ,0<x<\pi$
Solution:
$\tan ^{ -1 }{ \left( \sqrt { \frac { 1-cosx }{ 1+cosx } } \right) } ,0<x<\pi$
= $\tan ^{ -1 }{ \left[ \sqrt { \frac { { 2sin }^{ 2 }\frac { x }{ 2 } }{ { 2cos }^{ 2 }\frac { x }{ 2 } } } \right] }$
NCERT Solutions for Class 12 Maths Chapter 2 Inverse Trigonometric Functions 7

Question 8.
$\tan ^{ -1 }{ \left( \frac { cosx-sinx }{ cosx+sinx } \right) ,0<x<\pi }$
Solution:
$\tan ^{ -1 }{ \left( \frac { cosx-sinx }{ cosx+sinx } \right) ,0<x<\pi }$
Dividing numerator and denominator by cos x
NCERT Solutions for Class 12 Maths Chapter 2 Inverse Trigonometric Functions 8

Question 9.
$\tan ^{ -1 }{ \left( \frac { x }{ \sqrt { { a }^{ 2 }-{ x }^{ 2 } } } \right) ,\left| x \right| } <a$
Solution:
Let x = a sinθ
=> $\\ \frac { x }{ a }$ = sinθ
NCERT Solutions for Class 12 Maths Chapter 2 Inverse Trigonometric Functions 9

Question 10.
$\tan ^{ -1 }{ \left[ \frac { { 3a }^{ 2 }-{ x }^{ 3 } }{ { a }^{ 3 }-{ 3ax }^{ 2 } } \right] ,a>0;\frac { -a }{ \sqrt { 3 } } <x,<\frac { a }{ \sqrt { 3 } } }$
Solution:
Put x = a tanθ,
we get
NCERT Solutions for Class 12 Maths Chapter 2 Inverse Trigonometric Functions 10

Question 11.
Find the value of the following
$\tan ^{ -1 }{ \left[ 2cos\left( 2\sin ^{ -1 }{ \frac { 1 }{ 2 } } \right) \right] }$
Solution:
$\tan ^{ -1 }{ \left[ 2cos\left( 2\sin ^{ -1 }{ \frac { 1 }{ 2 } } \right) \right] }$
= $\tan ^{ -1 }{ \left[ 2cos2.\frac { \pi }{ 6 } \right] }$
NCERT Solutions for Class 12 Maths Chapter 2 Inverse Trigonometric Functions 11

Question 12.
cot[tan^-1 a + cot^-1 a]
Solution:
Given
cot[tan^-1 a + cot^-1 a]
= $cot\left( \tan ^{ -1 }{ a } +\tan ^{ -1 }{ \frac { 1 }{ a } } \right)$
NCERT Solutions for Class 12 Maths Chapter 2 Inverse Trigonometric Functions 12

Question 13.
$tan\frac { 1 }{ 2 } \left[ \sin ^{ -1 }{ \frac { 2x }{ 1+{ x }^{ 2 } } +\cos ^{ -1 }{ \frac { 1-{ y }^{ 2 } }{ 1+{ y }^{ 2 } } } } \right] \left| x \right| <1,y>0\quad and\quad xy<1$
Solution:
Putting x = tanθ
=> tan^-1 x = θ
NCERT Solutions for Class 12 Maths Chapter 2 Inverse Trigonometric Functions 13

Question 14.
If $sin\left( \sin ^{ -1 }{ \frac { 1 }{ 5 } } +\cos ^{ -1 }{ x } \right) =1$ then find the value of x
Solution:
$sin\left( \sin ^{ -1 }{ \frac { 1 }{ 5 } } +\cos ^{ -1 }{ x } \right) =sin\frac { \pi }{ 2 }$
NCERT Solutions for Class 12 Maths Chapter 2 Inverse Trigonometric Functions 14

Question 15.
If $\tan ^{ -1 }{ \frac { x-1 }{ x-2 } } +\tan ^{ -1 }{ \frac { x+1 }{ x+2 } } =\frac { \pi }{ 4 }$ then find the value of x
Solution:
L.H.S
$\tan ^{ -1 }{ \frac { x-1 }{ x-2 } } +\tan ^{ -1 }{ \frac { x+1 }{ x+2 } } =\frac { \pi }{ 4 }$
NCERT Solutions for Class 12 Maths Chapter 2 Inverse Trigonometric Functions 15

Question 16.
$\sin ^{ -1 }{ \left( sin\frac { 2\pi }{ 3 } \right) }$
Solution:
$\sin ^{ -1 }{ \left( sin\frac { 2\pi }{ 3 } \right) }$
= $\sin ^{ -1 }{ \left( sin\left( \pi -\frac { \pi }{ 3 } \right) \right) }$
= $\sin ^{ -1 }{ \left( sin\left( \frac { \pi }{ 3 } \right) \right) } =\frac { \pi }{ 3 }$

Question 17.
$\tan ^{ -1 }{ \left( tan\frac { 3\pi }{ 4 } \right) }$
Solution:
$\tan ^{ -1 }{ \left( tan\frac { 3\pi }{ 4 } \right) }$
= $\tan ^{ -1 }{ \left( sin\frac { 3\pi }{ 4 } \right) }$
= $\tan ^{ -1 }{ tan\left( \pi -\frac { \pi }{ 4 } \right) }$
NCERT Solutions for Class 12 Maths Chapter 2 Inverse Trigonometric Functions 17

Question 18.
$tan\left( \sin ^{ -1 }{ \frac { 3 }{ 5 } +\cot ^{ -1 }{ \frac { 3 }{ 2 } } } \right)$
Solution:
$tan\left( \sin ^{ -1 }{ \frac { 3 }{ 5 } +\cot ^{ -1 }{ \frac { 3 }{ 2 } } } \right)$
Let $\sin ^{ -1 }{ \frac { 3 }{ 5 } = } \theta$
sinθ = $\\ \frac { 3 }{ 5 }$
NCERT Solutions for Class 12 Maths Chapter 2 Inverse Trigonometric Functions 18

Question 19.
$\cos ^{ -1 }{ \left( cos\frac { 7\pi }{ 6 } \right) }$ is equal to
(a) $\frac { 7\pi }{ 6 }$
(b) $\frac { 5\pi }{ 6 }$
(c) $\frac { \pi }{ 5 }$
(d) $\frac { \pi }{ 6 }$
Solution:
$\cos ^{ -1 }{ \left( cos\frac { 7\pi }{ 6 } \right) }$
= $\cos ^{ -1 }{ cos\left( \pi +\frac { \pi }{ 6 } \right) }$
NCERT Solutions for Class 12 Maths Chapter 2 Inverse Trigonometric Functions 19

Question 20.
$sin\left[ \frac { \pi }{ 3 } -\sin ^{ -1 }{ \left( -\frac { 1 }{ 2 } \right) } \right]$ is equal to
(a) $\\ \frac { 1 }{ 2 }$
(b) $\\ \frac { 1 }{ 3 }$
(c) $\\ \frac { 1 }{ 4 }$
(d) 1
Solution:
$sin\left[ \frac { \pi }{ 3 } -\sin ^{ -1 }{ \left( -\frac { 1 }{ 2 } \right) } \right]$
NCERT Solutions for Class 12 Maths Chapter 2 Inverse Trigonometric Functions 20

Question 21.
$\tan ^{ -1 }{ \sqrt { 3 } -\cot ^{ -1 }{ \left( -\sqrt { 3 } \right) } }$ is equal to
(a) π
(b) $-\frac { \pi }{ 2 }$
(c) 0
(d) 2√3
Solution:
$\tan ^{ -1 }{ \sqrt { 3 } -\cot ^{ -1 }{ \left( -\sqrt { 3 } \right) } }$
NCERT Solutions for Class 12 Maths Chapter 2 Inverse Trigonometric Functions 21

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NCERT Solutions for Class 12 Maths Chapter 2 Inverse Trigonometric Functions Ex 2.2

NCERT Solutions for Class 12 Maths Chapter 2 Inverse Trigonometric Functions Ex 2.2

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