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

Get Free NCERT Solutions for Class 12 Maths Chapter 2 Inverse Trigonometric Functions Ex 2.2 PDF in Hindi and English Medium. Sets Class 12 Maths NCERT Solutions are extremely helpful while doing your homework. Inverse Trigonometric Functions Exercise 2.2 Class 12 Maths NCERT Solutions were prepared by Experienced LearnCBSE.in Teachers. Detailed answers of all the questions in Chapter 2 Class 12 Inverse Trigonometric Functions Ex 2.2 provided in NCERT Textbook.

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Topics and Sub Topics in Class 11 Maths Chapter 2 Inverse Trigonometric Functions Ex 2.2:

Section Name	Topic Name
2	Inverse Trigonometric Functions
2.1	Introduction
2.2	Basic Concepts
2.3	Properties of Inverse Trigonometric Functions

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

Ex 2.2 Class 12 Maths 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 $3\sin ^{ -1 }{ x=\sin ^{ -1 }{ (3x-4x^{ 3 });x\in \left[ -\frac { 1 }{ 2 } ,\frac { 1 }{ 2 } \right] } }$

Ex 2.2 Class 12 Maths 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.

Ex 2.2 Class 12 Maths 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

Ex 2.2 Class 12 Maths 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

Ex 2.2 Class 12 Maths 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

Ex 2.2 Class 12 Maths 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
vedantu class 12 maths Chapter 2 Inverse Trigonometric Functions 6

Ex 2.2 Class 12 Maths 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

Ex 2.2 Class 12 Maths 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

Ex 2.2 Class 12 Maths 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

Ex 2.2 Class 12 Maths 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

Ex 2.2 Class 12 Maths Question 11.
Find the value of the following
$\tan ^{ -1 }{ \left[ 2cos\left( 2\sin ^{ -1 }{ \frac { 1 }{ 2 } } \right) \right] }$
Solution:
$\tan ^{ -1 }{ \left[ 2cos\left( 2\sin ^{ -1 }{ \frac { 1 }{ 2 } } \right) \right] }$
= $\tan ^{ -1 }{ \left[ 2cos2.\frac { \pi }{ 6 } \right] }$
vedantu class 12 maths Chapter 2 Inverse Trigonometric Functions 11

Ex 2.2 Class 12 Maths 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

Ex 2.2 Class 12 Maths 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

Ex 2.2 Class 12 Maths 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

Ex 2.2 Class 12 Maths 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
$\tan ^{ -1 }{ \frac { x-1 }{ x-2 } } +\tan ^{ -1 }{ \frac { x+1 }{ x+2 } } =\frac { \pi }{ 4 }$
vedantu class 12 maths Chapter 2 Inverse Trigonometric Functions 15

Ex 2.2 Class 12 Maths Question 16.
$\sin ^{ -1 }{ \left( sin\frac { 2\pi }{ 3 } \right) }$
Solution:
$\sin ^{ -1 }{ \left( sin\frac { 2\pi }{ 3 } \right) }$
= $\sin ^{ -1 }{ \left( sin\left( \pi -\frac { \pi }{ 3 } \right) \right) }$
= $\sin ^{ -1 }{ \left( sin\left( \frac { \pi }{ 3 } \right) \right) } =\frac { \pi }{ 3 }$

Ex 2.2 Class 12 Maths 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

Ex 2.2 Class 12 Maths 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

Ex 2.2 Class 12 Maths 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

Ex 2.2 Class 12 Maths 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:
$sin\left[ \frac { \pi }{ 3 } -\sin ^{ -1 }{ \left( -\frac { 1 }{ 2 } \right) } \right]$
vedantu class 12 maths Chapter 2 Inverse Trigonometric Functions 20

Ex 2.2 Class 12 Maths 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