NCERT Solutions for Class 12 Maths Chapter 2 Exercise 2.2 (Ex 2.2)

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. 

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

• 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

• 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

Access Other Exercises of Class 12 Maths Chapter 2