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Class 12th Chapter -2 Inverse Trigonometric Functions | NCERT Maths Solution | NCERT Solution | Edugrown

It is important that you have a strong foundation in Maths Solutions. This can help you immensely in exams. Because there is no paper pattern set for the board exams. Every year there are different twists and turns in questions asked in board exams. While preparing for maths it becomes to go through every bit of detail. Thus, NCERT Solutions Class 12 Maths can help you lay a strong foundation. There solutions of each and every question given in the NCERT 12 Textbook in this article. There are crucial chapters like integration, algebra, and calculus which you can solve easily using these Class 12 Maths solutions. Using these NCERT Maths Class 12 Solutions can give you an edge over the other students.

NCERT Solutions for Class 12 Maths Chapter : 2 Inverse Trigonometric

Class 12 Maths Chapter 2 Inverse Trigonometric Ex 2.1

Ex 2.1 Class 12 Maths Question 1-10.
Find the principal values of the following:
(1) $\sin ^{ -1 }{ \left( -\frac { 1 }{ 2 } \right) }$

(2) $\cos ^{ -1 }{ \left( \frac { \sqrt { 3 } }{ 2 } \right) }$
(3) $\csc ^{ -1 }{ (2) }$
(4) $\tan ^{ -1 }{ \left( -\sqrt { 3 } \right) }$
(5) $\cos ^{ -1 }{ \left( -\frac { 1 }{ 2 } \right) }$
(6) $\tan ^{ -1 }{ (-1) }$
(7) $\sec ^{ -1 }{ \left( \frac { 2 }{ \sqrt { 3 } } \right) }$
(8) $\cot ^{ -1 }{ \left( \sqrt { 3 } \right) }$
(9) $\cos ^{ -1 }{ \left( -\frac { 1 }{ \sqrt { 2 } } \right) }$
(10) $\csc ^{ -1 }{ \left( -\sqrt { 2 } \right) }$
Solution:
(1) Let $\sin ^{ -1 }{ \left( -\frac { 1 }{ 2 } \right) }$ = y
∴ $sin\quad y=-\frac { 1 }{ 2 } =-sin\frac { \pi }{ 6 } =sin\left( -\frac { \pi }{ 6 } \right)$
the range of principal value of sin^-1 is

NCERT Solutions for Class 12 Maths Chapter 2 Inverse Trigonometric Functions Ex 2.1 Q1.1

NCERT Solutions for Class 12 Maths Chapter 2 Inverse Trigonometric Functions Ex 2.1 Q1.2

NCERT Solutions for Class 12 Maths Chapter 2 Inverse Trigonometric Functions Ex 2.1 Q1.3

NCERT Solutions for Class 12 Maths Chapter 2 Inverse Trigonometric Functions Ex 2.1 Q1.4

Ex 2.1 Class 12 Maths Question 11-12.
Find the principal values of the following:
(11) $\tan ^{ -1 }{ (1) } +\cos ^{ -1 }{ \left( -\frac { 1 }{ 2 } \right) } +\sin ^{ -1 }{ \left( -\frac { 1 }{ 2 } \right) }$
(12) $\cos ^{ -1 }{ \left( \frac { 1 }{ 2 } \right) } +2\sin ^{ -1 }{ \left( \frac { 1 }{ 2 } \right) }$
Solution:
(11) $\tan ^{ -1 }{ (1) } +\cos ^{ -1 }{ \left( -\frac { 1 }{ 2 } \right) } +\sin ^{ -1 }{ \left( -\frac { 1 }{ 2 } \right) }$
Now tan^-1 (1) = $\frac { \pi }{ 4 }$
∴the range of principal value branch of
NCERT Solutions for Class 12 Maths Chapter 2 Inverse Trigonometric Functions Ex 2.1 Q2.1

Ex 2.1 Class 12 Maths Question 13.
If sin^-1 x = y, then
(a) 0 ≤ y ≤ π
(b) $-\frac { \pi }{ 2 } \le y\le \frac { \pi }{ 2 }$
(c) 0 < y < π
(d) $-\frac { \pi }{ 2 } <y<\frac { \pi }{ 2 }$
Solution:
The range of principal value of sin is $\left[ -\frac { \pi }{ 2 } ,\frac { \pi }{ 2 } \right]$
∴ if sin^-1 x = y then
$-\frac { \pi }{ 2 } \le y\le \frac { \pi }{ 2 }$
Option (b) is correct

Ex 2.1 Class 12 Maths Question 14.
$\tan ^{ -1 }{ \sqrt { 3 } -\sec ^{ -1 }{ (-2) } }$ is equal to
(a) π
(b) $-\frac { \pi }{ 3 }$
(c) $\frac { \pi }{ 3 }$
(d) $\frac { 2\pi }{ 3 }$
Solution:
$\tan ^{ -1 }{ \sqrt { 3 } =\frac { \pi }{ 3 } ,\sec ^{ -1 }{ (-2) } } =\pi -\frac { \pi }{ 3 } =\frac { 2\pi }{ 3 }$
∴ Principal values of sec^-1 is [0,π] – $\left\{ \frac { \pi }{ 2 } \right\}$
$\tan ^{ -1 }{ \sqrt { 3 } - } \sec ^{ -1 }{ (-2) } =\frac { \pi }{ 3 } -\frac { 2\pi }{ 3 } =-\frac { \pi }{ 3 }$
Option (b) is correct

Class 12 Maths Chapter 2 Inverse Trigonometric 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 Ex 2.2 Q4.1

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 Ex 2.2 Q5.1

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
NCERT Solutions for Class 12 Maths Chapter 2 Inverse Trigonometric Functions Ex 2.2 Q6.1

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 Ex 2.2 Q7.1

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 Ex 2.2 Q8.1

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 Ex 2.2 Q9.1

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 Ex 2.2 Q10.1

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:
$\tan ^{ -1 }{ \left[ 2cos\left( 2\sin ^{ -1 }{ \frac { 1 }{ 2 } } \right) \right] }$
= $\tan ^{ -1 }{ \left[ 2cos2.\frac { \pi }{ 6 } \right] }$
NCERT Solutions for Class 12 Maths Chapter 2 Inverse Trigonometric Functions Ex 2.2 Q11.1

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 Ex 2.2 Q12.1

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 Ex 2.2 Q13.1

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 Ex 2.2 Q14.1

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
$\tan ^{ -1 }{ \frac { x-1 }{ x-2 } } +\tan ^{ -1 }{ \frac { x+1 }{ x+2 } } =\frac { \pi }{ 4 }$
NCERT Solutions for Class 12 Maths Chapter 2 Inverse Trigonometric Functions Ex 2.2 Q15.1

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 Ex 2.2 Q17.1

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 Ex 2.2 Q18.1

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 Ex 2.2 Q19.1

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:
$sin\left[ \frac { \pi }{ 3 } -\sin ^{ -1 }{ \left( -\frac { 1 }{ 2 } \right) } \right]$
NCERT Solutions for Class 12 Maths Chapter 2 Inverse Trigonometric Functions Ex 2.2 Q20.1

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 Ex 2.2 Q21.1