NCERT Solutions For Class 12 Maths Chapter 2 Inverse Trigonometric Functions Ex 2.2 (2023)

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

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 MathsQuestion 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 MathsQuestion 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 MathsQuestion 4.
\(2\tan ^{ -1 }{ \frac { 1 }{ 2 } } +\tan ^{ -1 }{ \frac { 1 }{ 7 } } =\tan ^{ -1 }{ \frac { 31 }{ 17 } } \)
Solution:
L.H.S =
\(2\tan ^{ -1 }{ \frac { 1 }{ 2 } } +\tan ^{ -1 }{ \frac { 1 }{ 7 } } \)

Ex 2.2 Class 12 MathsQuestion 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

Ex 2.2 Class 12 MathsQuestion 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

Ex 2.2 Class 12 MathsQuestion 7.
\(\tan ^{ -1 }{ \left( \sqrt { \frac { 1-cosx }{ 1+cosx } } \right) } ,0<x<\pi \)
Solution:
\(\tan ^{ -1 }{ \left( \sqrt { \frac { 1-cosx }{ 1+cosx } } \right) } ,0<x<\pi \)
= \(\tan ^{ -1 }{ \left[ \sqrt { \frac { { 2sin }^{ 2 }\frac { x }{ 2 } }{ { 2cos }^{ 2 }\frac { x }{ 2 } } } \right] } \)

Ex 2.2 Class 12 MathsQuestion 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

Ex 2.2 Class 12 MathsQuestion 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

Ex 2.2 Class 12 MathsQuestion 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

Ex 2.2 Class 12 MathsQuestion 11.
Find the value of the following
\(\tan ^{ -1 }{ \left[ 2cos\left( 2\sin ^{ -1 }{ \frac { 1 }{ 2 } } \right) \right] } \)
Solution:
\(\tan ^{ -1 }{ \left[ 2cos\left( 2\sin ^{ -1 }{ \frac { 1 }{ 2 } } \right) \right] } \)
= \(\tan ^{ -1 }{ \left[ 2cos2.\frac { \pi }{ 6 } \right] } \)

Ex 2.2 Class 12 MathsQuestion 12.
cot[tan^-1 a + cot^-1 a]
Solution:
Given
cot[tan^-1 a + cot^-1 a]
= \(cot\left( \tan ^{ -1 }{ a } +\tan ^{ -1 }{ \frac { 1 }{ a } } \right) \)

Ex 2.2 Class 12 MathsQuestion 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

Ex 2.2 Class 12 MathsQuestion 14.
If \(sin\left( \sin ^{ -1 }{ \frac { 1 }{ 5 } } +\cos ^{ -1 }{ x } \right) =1\) then find the value of x
Solution:
\(sin\left( \sin ^{ -1 }{ \frac { 1 }{ 5 } } +\cos ^{ -1 }{ x } \right) =sin\frac { \pi }{ 2 } \)

Ex 2.2 Class 12 MathsQuestion 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
\(\tan ^{ -1 }{ \frac { x-1 }{ x-2 } } +\tan ^{ -1 }{ \frac { x+1 }{ x+2 } } =\frac { \pi }{ 4 } \)

Ex 2.2 Class 12 MathsQuestion 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 MathsQuestion 17.
\(\tan ^{ -1 }{ \left( tan\frac { 3\pi }{ 4 } \right) } \)
Solution:
\(\tan ^{ -1 }{ \left( tan\frac { 3\pi }{ 4 } \right) } \)
= \(\tan ^{ -1 }{ \left( sin\frac { 3\pi }{ 4 } \right) } \)
= \(\tan ^{ -1 }{ tan\left( \pi -\frac { \pi }{ 4 } \right) } \)

Ex 2.2 Class 12 MathsQuestion 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 = \(\\ \frac { 3 }{ 5 } \)

Ex 2.2 Class 12 MathsQuestion 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:
\(\cos ^{ -1 }{ \left( cos\frac { 7\pi }{ 6 } \right) } \)
= \(\cos ^{ -1 }{ cos\left( \pi +\frac { \pi }{ 6 } \right) } \)

Ex 2.2 Class 12 MathsQuestion 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:
\(sin\left[ \frac { \pi }{ 3 } -\sin ^{ -1 }{ \left( -\frac { 1 }{ 2 } \right) } \right] \)

Ex 2.2 Class 12 MathsQuestion 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 Hindi Medium Ex 2.2

Class 12 Maths NCERT Solutions

• Chapter 1 Relations and Functions
• Chapter 2 Inverse Trigonometric Functions
• Chapter 3 Matrices
• Chapter 4 Determinants
• Chapter 5 Continuity and Differentiability
• Chapter 6 Application of Derivatives
• Chapter 7 Integrals Ex 7.1
• Chapter 8 Application of Integrals
• Chapter 9 Differential Equations
• Chapter 10 Vector Algebra
• Chapter 11 Three Dimensional Geometry
• Chapter 12 Linear Programming
• Chapter 13 Probability Ex 13.1