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NCERT Solutions for Class 12 Maths Chapter 2: Inverse Trigonometric Functions - Exercise 2.2

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NCERT Solutions for Class 12 Maths Chapter 2 Exercise 2.2

Free PDF download of NCERT Solutions for Class 12 Maths Chapter 2 Exercise 2.2 (Ex 2.2) and all chapter exercises at one place prepared by expert teacher as per NCERT (CBSE) books guidelines. Class 12 Maths Chapter 2 Inverse Trigonometric Functions Exercise 2.2 Questions with Solutions to help you to revise complete Syllabus and Score More marks. Register and get all exercise solutions in your emails.


Class:

NCERT Solutions for Class 12

Subject:

Class 12 Maths

Chapter Name:

Chapter 2 - Inverse Trigonometric Functions

Exercise:

Exercise - 2.2

Content-Type:

Text, Videos, Images and PDF Format

Academic Year:

2023-24

Medium:

English and Hindi

Available Materials:

  • Chapter Wise

  • Exercise Wise

Other Materials

  • Important Questions

  • Revision Notes

Competitive Exams after 12th Science

Topics Covered in the NCERT Solutions For Class 12 Maths Chapter 2

Introduction

Basic Concepts

Properties of Inverse Trigonometric Functions

Miscellaneous Q&A


Access NCERT Solutions for Class 12 Mathematics Chapter 2 – Inverse Trigonometric Functions

Refer to page 10 for Exercise 2.2 in the PDF. 

1. Prove $3{\sin ^{ - 1}}x = {\sin ^{ - 1}}\left( {3x - 4{x^3}} \right),x \in \left[ { - \dfrac{1}{2},\dfrac{1}{2}} \right]$

Ans: To Prove $3{\sin ^{ - 1}}x = {\sin ^{ - 1}}\left( {3x - 4{x^3}} \right)$, where $x \in \left[ { - \dfrac{1}{2},\dfrac{1}{2}} \right]$

Let $x = \sin \theta .$ Then, ${\sin ^{ - 1}}x = \theta $.

We have given that,

R.H.S ${\sin ^{ - 1}}\left( {3x - 4{x^3}} \right) = {\sin ^{ - 1}}\left( {3\sin \theta  - 4{{\sin }^3}\theta } \right)$

$ = {\sin ^{ - 1}}(\sin 3\theta )$

$ = 3\theta $

$ = 3{\sin ^{ - 1}}x = L.H.S$

Hence proved.

2. Prove $3{\cos ^{ - 1}}x = {\cos ^{ - 1}}\left( {4{x^3} - 3x} \right),x \in \left[ {\dfrac{1}{2},1} \right]$

Ans: To Prove $3{\cos ^{ - 1}}x = {\cos ^{ - 1}}\left( {4{x^3} - 3x} \right),x \in \left[ {\dfrac{1}{2},1} \right]$

Let $x = \cos \theta $. Then, ${\cos ^{ - 1}}x = \theta $

We have given that,

R.H S =${\cos ^{ - 1}}\left( {4{x^3} - 3x} \right)$

$ = {\cos ^{ - 1}}\left( {4{{\cos }^3}\theta  - 3\cos \theta } \right)$

$ = {\cos ^{ - 1}}(\cos 3\theta )$

$ = 3\theta $

$ = 3{\cos ^{ - 1}}x = L.H.S$

Hence proved.

3. Prove ${\tan ^{ - 1}}\left( {\dfrac{2}{{11}}} \right) + {\tan ^{ - 1}}\left( {\dfrac{7}{{24}}} \right) = {\tan ^{ - 1}}\left( {\dfrac{1}{2}} \right)$

Ans: We have to prove that: ${\tan ^{ - 1}}\dfrac{2}{{11}} + {\tan ^{ - 1}}\dfrac{7}{{24}} = {\tan ^{ - 1}}\dfrac{1}{2}$

Consider, L.H.S. ${\tan ^{ - 1}}\dfrac{2}{{11}} + {\tan ^{ - 1}}\dfrac{7}{{24}}$

$ = {\tan ^{ - 1}}\dfrac{{\dfrac{2}{{11}} + \dfrac{7}{{24}}}}{{1 - \left( {\dfrac{2}{{11}} \cdot \dfrac{7}{{24}}} \right)}}\quad \left[ {{{\tan }^{ - 1}}x + {{\tan }^{ - 1}}y = {{\tan }^{ - 1}}\dfrac{{x + y}}{{1 - xy}}} \right]$

$ = {\tan ^{ - 1}}\dfrac{{11 \times 24}}{{\dfrac{{11 \times 24 - 14}}{{11 \times 24}}}}$

$ = {\tan ^{ - 1}}\dfrac{{48 + 77}}{{264 - 14}}$

$ = {\tan ^{ - 1}}\left( {\dfrac{{125}}{{250}}} \right) = {\tan ^{ - 1}}\left( {\dfrac{1}{2}} \right) = $ R.H.S.

Hence proved.

4. Prove $2{\tan ^{ - 1}}\dfrac{1}{2} + {\tan ^{ - 1}}\dfrac{1}{7} = {\tan ^{ - 1}}\dfrac{{31}}{{17}}$

Ans:  To prove: $2{\tan ^{ - 1}}\dfrac{1}{2} + {\tan ^{ - 1}}\dfrac{1}{7} = {\tan ^{ - 1}}\dfrac{{31}}{{17}}$

${\text{L}}{\text{.H}}{\text{.S}} = 2{\tan ^{ - 1}}\dfrac{1}{2} + {\tan ^{ - 1}}\dfrac{1}{7}$ 

$= {\tan ^{ - 1}}\dfrac{{2 \cdot \dfrac{1}{2}}}{{1 - {{\left( {\dfrac{1}{2}} \right)}^2}}} + {\tan ^{ - 1}}\dfrac{1}{7}\left[ {2{{\tan }^{ - 1}}x = {{\tan }^{ - 1}}\dfrac{{2x}}{{1 - {x^2}}}} \right]$

$= {\tan ^{ - 1}}\dfrac{1}{{\left( {\dfrac{3}{4}} \right)}} + {\tan ^{ - 1}}\dfrac{1}{7}$

$= {\tan ^{ - 1}}\dfrac{4}{3} + {\tan ^{ - 1}}\dfrac{1}{7}$

$= {\tan ^{ - 1}}\dfrac{{\dfrac{4}{3} + \dfrac{1}{7}}}{{1 - \dfrac{4}{3} \cdot \dfrac{1}{7}}}\left[ {{{\tan }^{ - 1}}x + {{\tan }^{ - 1}}y = {{\tan }^{ - 1}}\dfrac{{x + y}}{{1 - xy}}} \right]$

$= {\tan ^{ - 1}}\left( {\dfrac{{\dfrac{{28 + 3}}{{21}}}}{{\dfrac{{21 - 4}}{{21}}}}} \right)$ 

$= {\tan ^{ - 1}}\dfrac{{31}}{{17}} = {\text{ R}}{\text{.H}}{\text{.S}}{\text{.}}$

5. Write the function in the simplest form: ${\tan ^{ - 1}}\dfrac{{\sqrt {1 + {x^2}}  - 1}}{x},x \ne 0$

Ans: ${\tan ^{ - 1}}\dfrac{{\sqrt {1 + {x^2}}  - 1}}{x}$

By putting $x = \tan \theta  \Rightarrow \theta  = {\tan ^{ - 1}}x$

$\therefore {\tan ^{ - 1}}\dfrac{{\sqrt {1 + {x^2}}  - 1}}{x} = {\tan ^{ - 1}}\dfrac{{\sqrt {1 + {{\tan }^2}\theta }  - 1}}{{\tan \theta }}$

$ = {\tan ^{ - 1}}\left( {\dfrac{{\sec \theta  - 1}}{{\tan \theta }}} \right) = {\tan ^{ - 1}}\left( {\dfrac{{1 - \cos \theta }}{{\sin \theta }}} \right)$

${\tan ^{ - 1}}\left( {\dfrac{{2{{\sin }^2}\left( {\dfrac{\theta }{2}} \right)}}{{2\sin \dfrac{\theta }{2}\cos \dfrac{\theta }{2}}}} \right)$

$ = {\tan ^{ - 1}}\left( {\tan \dfrac{\theta }{2}} \right) = \dfrac{\theta }{2} = \dfrac{1}{2}{\tan ^{ - 1}}x$

6. Write the function in the simplest form: ${\tan ^{ - 1}}\left( {\dfrac{1}{{\sqrt {{x^2} - 1} }}} \right),|x| > 1$

Ans: We have given that ${\tan ^{ - 1}}\dfrac{1}{{\sqrt {{x^2} - 1} }},|x| > 1$

By putting  $x = \operatorname{cosec} \theta  \Rightarrow \theta  = {\operatorname{cosec} ^{ - 1}}x$

$\therefore {\tan ^{ - 1}}\dfrac{1}{{\sqrt {{x^2} - 1} }}$

$ = {\tan ^{ - 1}}\dfrac{1}{{\sqrt {\cos e{c^2}\theta  - 1} }}$

$ = {\tan ^{ - 1}}\left( {\dfrac{1}{{\cot \theta }}} \right)$

$ = {\tan ^{ - 1}}(\tan \theta )$

$ = \theta $

$ = \cos e{c^{ - 1}}x$

$ = \dfrac{\pi }{2} - {\sec ^{ - 1}}x\quad \left[ {As,\cos e{c^{ - 1}}x + {{\sec }^{ - 1}}x = \dfrac{\pi }{2}} \right]$

7. Write the function in the simplest form: ${\tan ^{ - 1}}\left( {\sqrt {\dfrac{{1 - \cos x}}{{1 + \cos x}}} } \right),x < \pi $

Ans: Given that,${\tan ^{ - 1}}\left( {\sqrt {\dfrac{{1 - \cos x}}{{1 + \cos x}}} } \right),x < \pi $

${\tan ^{ - 1}}\left( {\sqrt {\dfrac{{1 - \cos x}}{{1 + \cos x}}} } \right)$

$ = {\tan ^{ - 1}}\left( {\sqrt {\dfrac{{2{{\sin }^2}\dfrac{x}{2}}}{{2{{\cos }^2}\dfrac{x}{2}}}} } \right)$

$ = {\tan ^{ - 1}}\left( {\dfrac{{\sin \dfrac{x}{2}}}{{\cos \dfrac{x}{2}}}} \right)$

$ = {\tan ^{ - 1}}\left( {\tan \dfrac{x}{2}} \right)$

$ = \dfrac{x}{2}$

8. Write the function in the simplest form: ${\tan ^{ - 1}}\left( {\dfrac{{\cos x - \sin x}}{{\cos x + \sin x}}} \right),0 < x < \pi $

Ans: We have given that, ${\tan ^{ - 1}}\left( {\dfrac{{\cos x - \sin x}}{{\cos x + \sin x}}} \right)$

$ = {\tan ^{ - 1}}\left( {\dfrac{{1 - \left( {\dfrac{{\sin x}}{{\cos x}}} \right)}}{{1 + \left( {\dfrac{{\sin x}}{{\cos x}}} \right)}}} \right)$

$ = {\tan ^{ - 1}}\left( {\dfrac{{1 - \tan x}}{{1 + \tan x}}} \right)$

$ = {\tan ^{ - 1}}(1) - {\tan ^{ - 1}}(\tan x)\quad \left[ {\because \quad \dfrac{{ - y}}{{x - xy}} = {{\tan }^{ - 1}}x - {{\tan }^{ - 1}}y} \right]$

$ = \dfrac{\pi }{4} - x$

9. Write the function in the simplest form: ${\tan ^{ - 1}}\dfrac{x}{{\sqrt {{a^2} - {x^2}} }},|x| < a$

Ans: We have given that,${\tan ^{ - 1}}\dfrac{x}{{\sqrt {{a^2} - {x^2}} }}$

Let’s consider, $x = a\sin \theta  \Rightarrow \dfrac{x}{a} = \sin \theta  \Rightarrow {\sin ^{ - 1}}\left( {\dfrac{x}{a}} \right)$

$\therefore {\tan ^{ - 1}}\dfrac{x}{{\sqrt {{a^2} - {x^2}} }}$

$ = {\tan ^{ - 1}}\left( {\dfrac{{a\sin \theta }}{{\sqrt {{a^2} - {a^2}{{\sin }^2}\theta } }}} \right)$

$ = {\tan ^{ - 1}}\left( {\dfrac{{a\sin \theta }}{{a\sqrt {1 - {{\sin }^2}\theta } }}} \right)$

$ = {\tan ^{ - 1}}\left( {\dfrac{{a\sin \theta }}{{a\cos \theta }}} \right)$

$ = {\tan ^{ - 1}}(\tan \theta ) = \theta  = {\sin ^{ - 1}}\dfrac{x}{a}$

10. Write the function in the simplest form: ${\tan ^{ - 1}}\left( {\dfrac{{3{a^2}x - {x^3}}}{{{a^3} - 3a{x^2}}}} \right),a > 0;\dfrac{{ - a}}{{\sqrt 3 }} \leqslant x \leqslant \dfrac{a}{{\sqrt 3 }}$

Ans: Consider, ${\tan ^{ - 1}}\left( {\dfrac{{3{a^2}x - {x^3}}}{{{a^3} - 3a{x^2}}}} \right)$

Let’s consider, $x = a\tan \theta  \Rightarrow \dfrac{x}{a} = \tan \theta  \Rightarrow \theta  = {\tan ^{ - 1}}\left( {\dfrac{x}{a}} \right)$

${\tan ^{ - 1}}\left( {\dfrac{{3{a^2}x - {x^3}}}{{{a^3} - 3a{x^2}}}} \right)$

$ = {\tan ^{ - 1}}\left( {\dfrac{{3{a^2} \cdot a\tan \theta  - {a^3}{{\tan }^3}\theta }}{{{a^3} - 3a \cdot {a^2}{{\tan }^2}\theta }}} \right)$

$ = {\tan ^{ - 1}}\left( {\dfrac{{3{a^3}\tan \theta  - {a^3}{{\tan }^3}\theta }}{{{a^3} - 3{a^3}{{\tan }^2}\theta }}} \right)$

$ = {\tan ^{ - 1}}(\tan 3\theta )$

$ = 3\theta $

$ = 3{\tan ^{ - 1}}\dfrac{x}{a}$

11. Find the value of ${\tan ^{ - 1}}\left[ {2\cos \left( {2{{\sin }^{ - 1}}\dfrac{1}{2}} \right)} \right]$

Ans: Let’s consider, ${\sin ^{ - 1}}\dfrac{1}{2} = x$

Then $\sin x = \dfrac{1}{2} = \sin \left( {\dfrac{\pi }{6}} \right)$

$\therefore {\sin ^{ - 1}}\dfrac{1}{2} = \dfrac{\pi }{6}$

$\therefore {\tan ^{ - 1}}\left[ {2\cos \left( {2{{\sin }^{ - 1}}\dfrac{1}{2}} \right)} \right]$

$ = {\tan ^{ - 1}}\left[ {2\cos \left( {2 \times \dfrac{\pi }{6}} \right)} \right]$

$ = {\tan ^{ - 1}}\left[ {2\cos \dfrac{\pi }{3}} \right]$

$ = {\tan ^{ - 1}}\left[ {2 \times \dfrac{1}{2}} \right]$

$ = {\tan ^{ - 1}}1 = \dfrac{\pi }{4}$

12. Find the value of $\cot \left( {{{\tan }^{ - 1}}a + {{\cot }^{ - 1}}a} \right)$

Ans: We have given that, $\cot \left( {{{\tan }^{ - 1}}a + {{\cot }^{ - 1}}a} \right)$

$ = \cot \left( {\dfrac{\pi }{2}} \right)\quad \left[ {\because \quad  - {{\cot }^{ - 1}}x = \dfrac{\pi }{2}} \right]$

$ = 0$

13. Find the value of $\tan \dfrac{1}{2}\left[ {{{\sin }^{ - 1}}\dfrac{{2x}}{{1 + {x^2}}} + {{\cos }^{ - 1}}\dfrac{{1 - {y^2}}}{{1 + {y^2}}}} \right],|x| < 1,y > 0$ and $xy < 1$

Ans: Let consider, $x = \tan \theta $.

Then, $\theta  = {\tan ^{ - 1}}x$.

$\therefore {\sin ^{ - 1}}\dfrac{{2x}}{{1 + {x^2}}} = {\sin ^{ - 1}}\left( {\dfrac{{2\tan \theta }}{{1 + {{\tan }^2}\theta }}} \right)$

$ = {\sin ^{ - 1}}(\sin 2\theta )$

$ = 2\theta $

$ = 2{\tan ^{ - 1}}x$

Let’s assume, $y = \tan \theta .$ Then, $\theta  = {\tan ^{ - 1}}y$.

$\therefore {\cos ^{ - 1}}\left( {\dfrac{{1 - {y^2}}}{{1 + {y^2}}}} \right)$

$ = {\cos ^{ - 1}}\left( {\dfrac{{1 - {{\tan }^2}\theta }}{{1 + {{\tan }^2}\theta }}} \right)$

$ = {\cos ^{ - 1}}(\cos 2\theta )$

$ = 2\theta  = 2{\tan ^{ - 1}}y$

$\therefore \tan \dfrac{1}{2}\left[ {{{\sin }^{ - 1}}\left( {\dfrac{{2x}}{{1 + {x^2}}}} \right) + {{\cos }^{ - 1}}\left( {\dfrac{{1 - {y^2}}}{{1 + {y^2}}}} \right)} \right]$

$ = \tan \dfrac{1}{2}\left[ {2{{\tan }^{ - 1}}x + 2{{\tan }^{ - 1}}y} \right]$

$\left[ {As,{{\tan }^{ - 1}}x + {{\tan }^{ - 1}}y = {{\tan }^{ - 1}}\dfrac{{x + y}}{{1 - xy}}} \right]$

$ = \tan \left[ {{{\tan }^{ - 1}}\left( {\dfrac{{x + y}}{{1 - xy}}} \right)} \right]$

$ = \dfrac{{x + y}}{{1 - xy}}$

14. If $\sin \left( {{{\sin }^{ - 1}}\dfrac{1}{5} + {{\cos }^{ - 1}}x} \right) = 1$, then find the value of ${\text{x}}$.

Ans: We have given that,$\sin \left( {{{\sin }^{ - 1}}\dfrac{1}{5} + {{\cos }^{ - 1}}x} \right) = 1$

$ \Rightarrow \sin \left( {{{\sin }^{ - 1}}\dfrac{1}{5}} \right)\cos \left( {{{\cos }^{ - 1}}x} \right) + \cos \left( {{{\sin }^{ - 1}}\dfrac{1}{5}} \right)\sin \left( {{{\cos }^{ - 1}}x} \right) = 1$

$[\because \quad B) = \sin A \cdot \cos B + \cos A \cdot \sin B]$

$ \Rightarrow \dfrac{1}{5} \cdot x + \cos \left( {{{\sin }^{ - 1}}\dfrac{1}{5}} \right)\sin \left( {{{\cos }^{ - 1}}x} \right) = 1$

$ \Rightarrow \dfrac{x}{5} + \cos \left( {{{\sin }^{ - 1}}\dfrac{1}{5}} \right)\sin \left( {{{\cos }^{ - 1}}x} \right) = 1$

Now, let consider ${\sin ^{ - 1}}\dfrac{1}{5} = y$

Therefore,

${\sin ^{ - 1}}\dfrac{1}{5} = y$

$\sin y = \dfrac{1}{5}$

$ \Rightarrow \cos y = \sqrt {1 - {{\left( {\dfrac{1}{5}} \right)}^2}}  = \dfrac{{2\sqrt 6 }}{5}$

$ \Rightarrow y = {\cos ^{ - 1}}\left( {\dfrac{{2\sqrt 6 }}{5}} \right)$

$\therefore {\sin ^{ - 1}}\dfrac{1}{5} = {\cos ^{ - 1}}\left( {\dfrac{{2\sqrt 6 }}{5}} \right)$

Let’s assume ${\cos ^{ - 1}}x = z$.

Then, $\cos z = x$

$ \Rightarrow \sin z = \sqrt {1 - {x^2}} $

$ \Rightarrow z = {\sin ^{ - 1}}\left( {\sqrt {1 - {x^2}} } \right)$

$\therefore {\cos ^{ - 1}}x = {\sin ^{ - 1}}\left( {\sqrt {1 - {x^2}} } \right)$... (3)

From equations  (1), (2) and (3) we have:

$\dfrac{x}{5} + \cos \left( {{{\cos }^{ - 1}}\dfrac{{2\sqrt 6 }}{5}} \right) \cdot \sin \left( {{{\sin }^{ - 1}}\sqrt {1 - {x^2}} } \right) = 1$

$ \Rightarrow \dfrac{x}{5} + \dfrac{{2\sqrt 6 }}{5} \cdot \sqrt {1 - {x^2}}  = 1$

$ \Rightarrow x + 2\sqrt 6 \sqrt {1 - {x^2}}  = 5$

$ \Rightarrow 2\sqrt 6 \sqrt {1 - {x^2}}  = 5 - x$

Squaring on both sides, we get:

${(2\sqrt 6 )^2}\left( {1 - {x^2}} \right) = 25 + {x^2} - 10x$

$ \Rightarrow (4 \times 6)\left( {1 - {x^2}} \right) = 25 + {x^2} - 10x$

$ \Rightarrow 24 - 24{x^2} = 25 + {x^2} - 10x$

$ \Rightarrow 25{x^2} - 10x + 1 = 0$

$ \Rightarrow {(5x - 1)^2} = 0$

$ \Rightarrow (5x - 1) = 0$

$ \Rightarrow x = \dfrac{1}{5}$

Therefore, the value of $x$ is $\dfrac{1}{5}$.

15. If ${\tan ^{ - 1}}\dfrac{{x - 1}}{{x - 2}} + {\tan ^{ - 1}}\dfrac{{x + 1}}{{x + 2}} = \dfrac{\pi }{4}$, then find the value of ${\text{x}}$.

Ans: We have given that, ${\tan ^{ - 1}}\dfrac{{x - 1}}{{x - 2}} + {\tan ^{ - 1}}\dfrac{{x + 1}}{{x + 2}} = \dfrac{\pi }{4}$

$ \Rightarrow {\tan ^{ - 1}}\left[ {\dfrac{{\dfrac{{x - 1}}{{x - 2}} + \dfrac{{x + 1}}{{x + 2}}}}{{1 - \left( {\dfrac{{x - 1}}{{x + 2}}} \right)\left( {\dfrac{{x + 1}}{{x + 2}}} \right)}}} \right] = \dfrac{\pi }{4}$

$\left[ {As,{{\tan }^{ - 1}}x + {{\tan }^{ - 1}}y = {{\tan }^{ - 1}}\dfrac{{x + y}}{{1 - xy}}} \right]$

$ \Rightarrow {\tan ^{ - 1}}\left[ {\dfrac{{(x - 1)(x + 2) + (x + 1)(x - 2)}}{{(x + 2)(x - 2) - (x - 1)(x + 1)}}} \right] = \dfrac{\pi }{4}$

$ \Rightarrow {\tan ^{ - 1}}\left[ {\dfrac{{{x^2} + x - 2 + {x^2} - x - 2}}{{{x^2} - 4 - {x^2} + 1}}} \right] = \dfrac{\pi }{4}$

$ \Rightarrow {\tan ^{ - 1}}\left[ {\dfrac{{2{x^2} - 4}}{{ - 3}}} \right] = \dfrac{\pi }{4}$

$ \Rightarrow \tan \left[ {{{\tan }^{ - 1}}\dfrac{{4 - 2{x^2}}}{3}} \right] = \tan \dfrac{\pi }{4}$

$ \Rightarrow \dfrac{{4 - 2{x^2}}}{3} = 1$

$ \Rightarrow 4 - 2{x^2} = 3$

$ \Rightarrow 2{x^2} = 4 - 3 = 1$

$ \Rightarrow x =  \pm \dfrac{1}{{\sqrt 2 }}$

Therefore, the value of ${\text{x}}$ is $ \pm \dfrac{1}{{\sqrt 2 }}$.

16. Find the values of ${\sin ^{ - 1}}\left( {\sin \dfrac{{2\pi }}{3}} \right)$

Ans: Let’s consider, ${\sin ^{ - 1}}\left( {\sin \dfrac{{2\pi }}{3}} \right)$

As we know that ${\sin ^{ - 1}}(\sin x) = x$

If $x \in \left[ { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right]$, which is the principal value branch of ${\sin ^{ - 1}}x$.

Here, $\dfrac{{2\pi }}{3} \notin \left[ {\dfrac{{ - \pi }}{2},\dfrac{\pi }{2}} \right]$

Now, ${\sin ^{ - 1}}\left( {\sin \dfrac{{2\pi }}{3}} \right)$ can be written as:

${\sin ^{ - 1}}\left( {\sin \dfrac{{2\pi }}{3}} \right)$

$ = {\sin ^{ - 1}}\left[ {\sin \left( {\pi  - \dfrac{{2\pi }}{3}} \right)} \right]$

$ = {\sin ^{ - 1}}\left( {\sin \dfrac{\pi }{3}} \right)$, where $\dfrac{\pi }{3} \in \left[ {\dfrac{{ - \pi }}{2},\dfrac{\pi }{2}} \right]$

$\therefore {\sin ^{ - 1}}\left( {\sin \dfrac{{2\pi }}{3}} \right) = {\sin ^{ - 1}}\left[ {\sin \dfrac{\pi }{3}} \right] = \dfrac{\pi }{3}$

17. Find the values of ${\tan ^{ - 1}}\left( {\tan \dfrac{{3\pi }}{4}} \right)$

Ans: Let’s Consider, ${\tan ^{ - 1}}\left( {\tan \dfrac{{3\pi }}{4}} \right)$

As we know that ${\tan ^{ - 1}}(\tan x) = x$

If $x \in \left( { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right)$, which is the principal value branch of ${\tan ^{ - 1}}x$. Here, $\dfrac{{3\pi }}{4} \notin \left( { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right)$.

Now, ${\tan ^{ - 1}}\left( {\tan \dfrac{{3\pi }}{4}} \right)$ can be written as:

${\tan ^{ - 1}}\left( {\tan \dfrac{{3\pi }}{4}} \right) = {\tan ^{ - 1}}\left[ { - \tan \left( {\dfrac{{ - 3\pi }}{4}} \right)} \right] = {\tan ^{ - 1}}\left[ { - \tan \left( {\pi  - \dfrac{\pi }{4}} \right)} \right]$

${\tan ^{ - 1}}\left( { - \tan \dfrac{\pi }{4}} \right) = {\tan ^{ - 1}}\left[ {\tan \left( {\dfrac{{ - \pi }}{4}} \right)} \right]$ where $ - \dfrac{\pi }{4} \in \left( { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right)$

$\therefore {\tan ^{ - 1}}\left( {\tan \dfrac{{3\pi }}{4}} \right) = {\tan ^{ - 1}}\left[ {\tan \left( {\dfrac{{ - \pi }}{4}} \right)} \right] = \dfrac{{ - \pi }}{4}$

18. Find the values of $\tan \left( {{{\sin }^{ - 1}}\dfrac{3}{5} + {{\cot }^{ - 1}}\dfrac{3}{2}} \right)$

Ans: Let’s consider, ${\sin ^{ - 1}}\dfrac{3}{5} = x$.

Then, $\sin x = \dfrac{3}{5}$

$ \Rightarrow \cos x = \sqrt {1 - {{\sin }^2}x}  = \dfrac{4}{5}$

$ \Rightarrow \sec x = \dfrac{5}{4}$

$\therefore \tan x = \sqrt {{{\sec }^2}x - 1}  = \sqrt {\dfrac{{25}}{{16}} - 1}  = \dfrac{3}{4}$

$\therefore x = {\tan ^{ - 1}}\dfrac{3}{4}$

$\therefore {\sin ^{ - 1}}\dfrac{3}{5} = {\tan ^{ - 1}}\dfrac{3}{4}\quad  \ldots (i)$

Therefore, $\tan \left( {{{\sin }^{ - 1}}\dfrac{3}{5} + {{\cot }^{ - 1}}\dfrac{3}{2}} \right)$

$ = \tan \left( {{{\tan }^{ - 1}}\dfrac{3}{4} + {{\tan }^{ - 1}}\dfrac{2}{3}} \right)\quad $ [Using (i) and (ii)]

$ = \tan \left[ {{{\tan }^{ - 1}}\left( {\dfrac{{\dfrac{3}{4} + \dfrac{2}{3}}}{{1 - \dfrac{3}{4} \cdot \dfrac{2}{3}}}} \right)} \right]$

$\left[ {As,{{\tan }^{ - 1}}x + {{\tan }^{ - 1}}y = {{\tan }^{ - 1}}\dfrac{{x + y}}{{1 - xy}}} \right]$

$ = \tan \left( {{{\tan }^{ - 1}}\dfrac{{9 + 8}}{{12 - 6}}} \right)$

$ = \tan \left( {{{\tan }^{ - 1}}\dfrac{{17}}{6}} \right) = \dfrac{{17}}{6}$

19. Find the values of ${\cos ^{ - 1}}\left( {\cos \dfrac{{7\pi }}{6}} \right)$ is equal to

(A) $\dfrac{{7\pi }}{6}$

(B) $\dfrac{{5\pi }}{6}$

(C) $\dfrac{\pi }{3}$

(D) $\dfrac{\pi }{6}$

Ans: We know that ${\cos ^{ - 1}}(\cos x) = x$ if $x \in [0,\pi ]$, which is the principal value branch of ${\cos ^{ - 1}}x$. Here, $\dfrac{{7\pi }}{6} \notin [0,\pi ]$.

Now, ${\cos ^{ - 1}}\left( {\cos \dfrac{{7\pi }}{6}} \right)$ can be written as:

${\cos ^{ - 1}}\left( {\cos \dfrac{{7\pi }}{6}} \right) = {\cos ^{ - 1}}\left( {\cos \dfrac{{ - 7\pi }}{6}} \right) = {\cos ^{ - 1}}\left[ {\cos \left( {2\pi  - \dfrac{{7\pi }}{6}} \right)} \right]\quad [\because $

$ + x) = \cos x]$

$\therefore {\cos ^{ - 1}}\left( {\cos \dfrac{{7\pi }}{6}} \right) = {\cos ^{ - 1}}\left( {\cos \dfrac{{5\pi }}{6}} \right) = \dfrac{{5\pi }}{6}$

The correct answer is ${\text{B}}$.

20. Find the values of $\sin \left( {\dfrac{\pi }{3} - {{\sin }^{ - 1}}\left( { - \dfrac{1}{2}} \right)} \right)$ is equal to

(A) $\dfrac{1}{2}$

(B) $\dfrac{1}{3}$

(C) $\dfrac{1}{4}$

(D) 1

Ans: Let’s consider, ${\sin ^{ - 1}}\left( {\dfrac{{ - 1}}{2}} \right) = x$.

Then, $\sin x = \dfrac{{ - 1}}{2} =  - \sin \dfrac{\pi }{6} = \sin \left( {\dfrac{{ - \pi }}{6}} \right)$.

As we know that the range of the principal value branch of ${\sin ^{ - 1}}$ is $\left[ {\dfrac{{ - \pi }}{2},\dfrac{\pi }{2}} \right]$. ${\sin ^{ - 1}}\left( {\dfrac{{ - 1}}{2}} \right) = \dfrac{\pi }{6}$

$\therefore \sin \left( {\dfrac{\pi }{3} - {{\sin }^{ - 1}}\left( {\dfrac{{ - 1}}{2}} \right)} \right) = \sin \left( {\dfrac{\pi }{3} + \dfrac{\pi }{6}} \right) = \sin \left( {\dfrac{{3\pi }}{6}} \right) = \sin \left( {\dfrac{\pi }{2}} \right) = 1$

The correct answer is ${\text{D}}$.

21. $\tan ^{-1}(\sqrt{3})-\cot ^{-1}(-\sqrt{3})$ is equal to

(A) $\pi$

(B) $-\dfrac{\pi}{2}$

(C) 0

(D) $2 \sqrt{3}$

Ans:

Suppose that,

$A=\tan ^{-1}(\sqrt{3})-\cot ^{-1}(-\sqrt{3})$

As we know that,

$\cot ^{-1}(-x)=\pi-\cot ^{-1}(x)$

So,

$\cot ^{-1}(-\sqrt{3})=\pi-\cot ^{-1}(\sqrt{3})$

Put the value in the equation (i),

$A=\tan ^{-1}(\sqrt{3})-\pi+\cot ^{-1}(\sqrt{3})$

As we know that,

$\cot ^{-1} x=\dfrac{\pi}{2}-\tan ^{-1} x$

So,

$\cot ^{-1}(\sqrt{3})=\dfrac{\pi}{2}-\tan ^{-1}(\sqrt{3})$

Put the value in the equation (ii),

$A=\tan ^{-1}(\sqrt{3})-\pi+\dfrac{\pi}{2}-\tan ^{-1}(\sqrt{3}) $

$A=-\dfrac{\pi}{2}$

Hence, the correct option is B.

Important Formulas/Properties covered in the NCERT Solutions for Class 12 Maths Chapter 2

Some of important theorems/properties of inverse trigonometric functions are given below. These theorems/properties help to convert one function into another. 


Function

Domain

Range

Sin-1 x

\[\mathrm{\left [ -1,1 \right ]}\]

\[\mathrm{\left \{ -\frac{\pi }{2}, \frac{\pi }{2}\right \}}\]

Cos-1 x

\[\mathrm{\left [ -1,1 \right ]}\]

\[\mathrm{\left \{ 0,\pi  \right \}}\]

Tan-1 x

R

\[\mathrm{\left \{ -\frac{\pi }{2}, \frac{\pi }{2}\right \}}\]

Cosec-1 x

\[\mathrm{x\geq 1\: or\: x\leq -1}\]

\[\mathrm{[0,\frac{\pi }{2})(\frac{\pi }{2},\pi ]}\]

Sec-1 x

\[\mathrm{x\geq 1\: or\: x\leq -1}\]

\[\mathrm{[0,\frac{\pi }{2})(\frac{\pi }{2},\pi ]}\]

Cot-1 x

R

\[\mathrm{\left \{ 0,\pi  \right \}}\]


Complimentary Functions Related to Inverse Trignomentric Function

  1. Sin-1 x + Cos-1 x = \[\mathrm{\frac{\pi }{2}}\]

  2. Cosec-1 x + Sec-1 x = \[\mathrm{\frac{\pi }{2}}\] 

  3. Tan-1 x + Cot-1 x =  \[\mathrm{\frac{\pi }{2}}\]

  4. Tan-1 x + Tan-1 y = \[\mathrm{\tan ^{-1}\frac{x+y}{1-xy}\: xy< 1}\]

  5. Tan-1 x - Tan-1 y = \[\mathrm{\tan ^{-1}\frac{x-y}{1+xy}\: xy< -1}\]


NCERT Solutions for Class 12 Maths Chapter 2 Exercise 2.2

Opting for the NCERT solutions for Ex 2.2 Class 12 Maths is considered as the best option for the CBSE students when it comes to exam preparation. This chapter consists of many exercises. Out of which we have provided the Exercise 2.2 Class 12 Maths NCERT solutions on this page in PDF format. You can download this solution as per your convenience or you can study it directly from our website/ app online.

Vedantu in-house subject matter experts have solved the problems/ questions from the exercise with the utmost care and by following all the guidelines by CBSE. Class 12 students who are thorough with all the concepts from the Maths textbook and quite well-versed with all the problems from the exercises given in it, then any student can easily score the highest possible marks in the final exam. With the help of this Class 12 Maths Chapter 2 Exercise 2.2 solutions, students can easily understand the pattern of questions that can be asked in the exam from this chapter and also learn the marks weightage of the chapter. So that they can prepare themselves accordingly for the final exam.

Besides these NCERT solutions for Class 12 Maths Chapter 2 Exercise 2.2, there are plenty of exercises in this chapter which contain innumerable questions as well. All these questions are solved/answered by our in-house subject experts as mentioned earlier. Hence all of these are bound to be of superior quality and anyone can refer to these during the time of exam preparation. In order to score the best possible marks in the class, it is really important to understand all the concepts of the textbooks and solve the problems from the exercises given next to it. 

Do not delay any more. Download the NCERT solutions for Class 12 Maths Chapter 2 Exercise 2.2 from Vedantu website now for better exam preparation. If you have the Vedantu app in your phone, you can download the same through the app as well. The best part of these solutions is these can be accessed both online and offline as well.


Chapter wise NCERT Solutions for Class 12 Maths


Access Other Exercises of Class 12 Maths Chapter 2

Chapter 2 Inverse Trigonometric Functions Exercises in PDF Format

Exercise 2.1

14 Questions and Solutions

FAQs on NCERT Solutions for Class 12 Maths Chapter 2: Inverse Trigonometric Functions - Exercise 2.2

1. For the tests, which examples and questions from class 12th maths exercise 2.2 are the most crucial?

Exercise 2.2 in math class 12 comprises 21 questions and 6 examples (Examples 3, 4, 5, 6, 7, 8). It is imperative to complete all of the exercise's examples and questions. But because these examples and questions have already been asked in the tests numerous times, examples 3, 5, and 6 and questions 1, 2, 3, 4, 5, 6, 7, 8, 10, 12, 13, 14, 15, 18, 19, and 21 of exercise 2.2 of class 12th math are most crucial.

2. How many days are required to finish Math Exercise 2.2 for grade 12?

Students will need two to three days to do exercise 1.2 of class 12 math if they can devote two hours every day to it. This time is a rough estimate. Because no two students work at the same rate or with the same efficiency, this time may vary.

3. How can you simplify NCERT Solutions for Class 12 Maths Chapter 2 Exercise 2.2?

It will be simpler for the pupils to learn if they comprehend it and practise it. They will also receive full practise with the questions and the procedure from Vedantu ncert solutions for class 12th exercise 1.2.   Students will benefit greatly from asking the same questions over.

4. Is exercise 2.2 from Maths class 12 difficult for the test?

Exercise 2.2 in math class 12 is neither simple nor difficult. Because some of the examples and questions in this exercise are simple and some are difficult, it falls somewhere in the middle of easy and tough. The amount of difficulty for any topic or question does, however, vary from child to child. Therefore, whether or not exercise 2.2 of math class 12 is difficult depends on the students. Some youngsters find it challenging, some find it simple, and yet others find it somewhere in the centre.

5. What is exercise 2.2 based upon?

In class 11 trigonometry, questions based on sin 2x, sin 3x, cos 3x,   etc. are found in exercise 2.2. It also has questions that use square roots to simplify inverse trigonometric functions. The answers to the trigonometric function-related questions 14 and 15 are listed below. Once the function has been solved, we should check the outcome to see if it satisfies the equation before deciding on the final response.