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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 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.

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

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

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