Exercise 2.2 (Revised) - Chapter 2 Inverse Trigonometry class 12 ncert solutions Maths - SaraNextGen [2024-2025]

Updated On August 15, 2024

By SaraNextGen

NCERT Class 12 Maths Solutions: Chapter 2 - Inverse Trigonometry

Ex 2.2 Question 1:

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

Answer.

We know that: $\sin 3 \theta=3 \sin \theta-4 \sin ^3 \theta$
Putting $\sin \theta=x$
$
\begin{aligned}
& \Rightarrow \theta=\sin ^{-1} x \\
& \therefore \sin 3 \theta=3 \sin \theta-4 \sin ^3 \theta \\
& \Rightarrow \sin 3 \theta=3 x-4 x^3 \\
& \Rightarrow 3 \theta=\sin ^{-1}\left(3 x-4 x^3\right)
\end{aligned}
$

Putting $\theta=\sin ^{-1} x$,
$
3 \sin ^{-1} x=\sin ^{-1}\left(3 x-4 x^3\right)
$

Proved.

Ex 2.2 Question 2:

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

Answer.

We know that: $\cos 3 \theta=4 \cos ^3 \theta-3 \cos \theta$
Putting $\cos \theta=x$
$
\begin{aligned}
& \Rightarrow \theta=\cos ^{-1} x \\
& \therefore \cos 3 \theta=4 x^3-3 x \\
& \Rightarrow 3 \theta=\cos ^{-1}\left(3 x-4 x^3\right)
\end{aligned}
$

Putting $\theta=\cos ^{-1} x$,
$
3 \cos ^{-1} x=\cos ^{-1}\left(4 x^3-3 x\right) \text { Proved. }
$

Ex 2.2 Question 3:

Write the following functions in the simplest form:
$\tan ^{-1} \frac{\sqrt{1+x^2}-1}{x}, x \neq 0$

Answer.

Putting $x=\tan \theta$ so that $\theta=\tan ^{-1} x$
$
\Rightarrow \tan ^{-1} \frac{\sqrt{1+x^2}-1}{x}
$

$\begin{aligned}
& =\tan ^{-1} \frac{\sqrt{1+\tan ^2 \theta}-1}{\tan \theta} \\
& =\tan ^{-1} \frac{\sec \theta-1}{\tan \theta} \\
& =\tan ^{-1}\left(\frac{\frac{1}{\cos \theta}-1}{\frac{\sin \theta}{\cos \theta}}\right) \\
& =\tan ^{-1}\left(\frac{1-\cos \theta}{\sin \theta}\right) \\
& =\tan ^{-1}\left(\frac{2 \sin ^2 \frac{\theta}{2}}{2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}}\right) \\
& =\tan ^{-1}\left(\tan \frac{\theta}{2}\right) \\
& =\frac{\theta}{2}=\frac{1}{2} \tan ^{-1} x
\end{aligned}$

Ex 2.2 Question 4:

$\begin{aligned}
& \text {} \tan ^{-1} \sqrt{\frac{1-\cos x}{1+\cos x}}, x<\pi \\
& \text { Answer. } \tan ^{-1} \sqrt{\frac{1-\cos x}{1+\cos x}} \\
& =\tan ^{-1} \sqrt{\frac{2 \sin ^2 \frac{x}{2}}{2 \cos ^2 \frac{x}{2}}} \\
& =\tan ^{-1} \tan \frac{x}{2} \\
& =\frac{x}{2}
\end{aligned}$

Ex 2.2 Question 5:

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

Answer:

$\tan ^{-1}\left(\frac{\cos x-\sin x}{\cos x+\sin x}\right)$
Dividing the numerator and denominator by $\cos x$,
$
\begin{aligned}
& =\tan ^{-1}\left(\frac{1-\tan x}{1+\tan x}\right) \\
& =\tan ^{-1}\left(\frac{\tan \frac{\pi}{4}-\tan x}{1+\tan \frac{\pi}{4} \tan x}\right) \\
& =\tan ^{-1} \tan \left(\frac{\pi}{4}-x\right) \quad\left[\tan (x-y)=\frac{\tan x-\tan y}{1+\tan x \cdot \tan y}\right] \\
& =\frac{\pi}{4}-x
\end{aligned}
$

Ex 2.2 Question 6:

$\tan ^{-1} \frac{x}{\sqrt{a^2-x^2}},|x|$

Answer.

Putting $x=a \sin \theta$ so that $\theta=\sin ^{-1} \frac{x}{a}$
$
\begin{aligned}
& \Rightarrow \tan ^{-1}\left(\frac{a \sin \theta}{\sqrt{a^2-a^2 \sin ^2 \theta}}\right) \\
& =\tan ^{-1}\left(\frac{a \sin \theta}{\sqrt{a^2\left(1-\sin ^2 \theta\right)}}\right) \\
& =\tan ^{-1}\left(\frac{a \sin \theta}{\sqrt{a^2 \cos ^2 \theta}}\right)
\end{aligned}
$

$\begin{aligned}
& =\tan ^{-1}\left(\frac{a \sin \theta}{a \cos \theta}\right) \\
& =\tan ^{-1} \tan \theta \\
& =\theta=\sin ^{-1} \frac{x}{a}
\end{aligned}$

Ex 2.2 Question 7:

$\tan ^{-1}\left(\frac{3 a^2 x-x^3}{a^3-3 a x^2}\right), a>0,\left(-\frac{a}{\sqrt{3}} \leq x \leq \frac{a}{\sqrt{3}}\right)$

Answer.

$\tan ^{-1}\left(\frac{3 a^2 x-x^3}{a^3-3 a x^2}\right)$
$=\tan ^{-1}\left(\frac{3\left(\frac{x}{a}\right)-\left(\frac{x}{a}\right)^3}{1-3\left(\frac{x}{a}\right)^2}\right)$ [Dividing numerator and denominator by $a^3$ ]
Putting $\frac{x}{a}=\tan \theta$ so that $\theta=\tan ^{-1} \frac{x}{a}$
$
\begin{aligned}
& =\tan ^{-1}\left(\frac{3 \tan \theta-\tan ^3 \theta}{1-3 \tan ^2 \theta}\right) \\
& =\tan ^{-1} \tan 3 \theta \\
& =3 \theta=3 \tan ^{-1} \frac{x}{a}
\end{aligned}
$

Ex 2.2 Question 8:

Find the values of each of the following:
$\tan ^{-1}\left[2 \cos \left(2 \sin ^{-1} \frac{1}{2}\right)\right]$

$\begin{aligned}
& \text { Ans. } \tan ^{-1}\left[2 \cos \left(2 \sin ^{-1} \frac{1}{2}\right)\right] \\
& =\tan ^{-1}\left[2 \cos \left(2 \sin ^{-1} \sin \frac{\pi}{6}\right)\right] \\
& =\tan ^{-1}\left[2 \cos \left(2 \times \frac{\pi}{6}\right)\right] \\
& =\tan ^{-1}\left[2 \cos \frac{\pi}{3}\right] \\
& =\tan ^{-1}\left[2 \times \frac{1}{2}\right] \\
& =\tan ^{-1} 1 \\
& =\tan ^{-1} \tan \frac{\pi}{4}=\frac{\pi}{4}
\end{aligned}$

Ex 2.2 Question 9

$\tan \frac{1}{2}\left[\sin ^{-1} \frac{2 x}{1+x^2}+\cos ^{-1} \frac{1-y^2}{1+y^2}\right],|x|<1, y>0$ and $x y<1$

Answer.

Putting $x=\tan \theta$ and $y=\tan \phi$
$
\therefore \tan \frac{1}{2}\left[\sin ^{-1} \frac{2 x}{1+x^2}+\cos ^{-1} \frac{1-y^2}{1+y^2}\right]
$

$\begin{aligned}
& =\tan \frac{1}{2}\left[\sin ^{-1} \frac{2 \tan \theta}{1+\tan ^2 \theta}+\cos ^{-1} \frac{1-\tan ^2 \phi}{1+\tan ^2 \phi}\right] \\
& =\tan \frac{1}{2}\left[\sin ^{-1} \sin 2 \theta+\cos ^{-1} \cos 2 \phi\right] \\
& =\tan \frac{1}{2}[2 \theta+2 \phi] \\
& =\tan [\theta+\phi] \\
& =\frac{\tan \theta+\tan \phi}{1-\tan \theta \tan \phi} \\
& =\frac{x+y}{1-x y}
\end{aligned}$

Ex 2.2 Question 10

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

Answer:

$\sin ^{-1}\left(\sin \frac{2 \pi}{3}\right)$
$
\begin{aligned}
& =\sin ^{-1}\left(\sin \frac{3 \pi-\pi}{3}\right) \\
& =\sin ^{-1}\left[\sin \left(\pi-\frac{\pi}{3}\right)\right] \\
& =\sin ^{-1} \sin \frac{\pi}{3} \\
& =\frac{\pi}{3}
\end{aligned}
$

Ex 2.2 Question 11:

$\tan ^{-1}\left(\tan \frac{3 \pi}{4}\right)$

Answer.

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

Ex 2.2 Question 12:

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

Answer.

Putting $\sin ^{-1} \frac{3}{5}=x$ and $\cot ^{-1} \frac{3}{2}=y$ so that $\sin x=\frac{3}{5}$ and $\cot y=\frac{3}{2}$
Now, $\cos x=\sqrt{1-\sin ^2 x}=\sqrt{1-\frac{9}{25}}=\sqrt{\frac{16}{25}}=\frac{4}{5}$
And $\tan x=\frac{\sin x}{\cos x}=\frac{3}{4}$
and $\tan y=\frac{1}{\cot y}=\frac{2}{3}$
$\therefore \tan \left(\sin ^{-1} \frac{3}{5}+\cot ^{-1} \frac{3}{2}\right)$
$=\tan (x+y)$
$=\frac{\tan x+\tan y}{1-\tan x \tan y}=\frac{\frac{3}{4}+\frac{2}{3}}{1-\frac{3}{4} \times \frac{2}{3}}$
$=\frac{\frac{17}{12}}{\frac{1}{2}}=\frac{17}{6}$

Ex 2.2 Question 13:

$\cos ^{-1}\left(\cos \frac{7 \pi}{6}\right)$ is equal to:
(A) $\frac{7 \pi}{6}$
(B) $\frac{5 \pi}{6}$

(C) $\frac{\pi}{3}$
(D) $\frac{\pi}{6}$

Answer:

$\cos ^{-1}\left(\cos \frac{7 \pi}{6}\right)$
$=\cos ^{-1}\left[\cos \left(2 \pi-\frac{7 \pi}{6}\right)\right][\because \cos (2 \pi-\theta)=\cos \theta]$
$=2 \pi-\frac{7 \pi}{6}=\frac{12 \pi-7 \pi}{6}=\frac{5 \pi}{6}$
Therefore, option (B) is correct.

Ex 2.2 Question 14:

$\sin \left[\frac{\pi}{3}-\sin ^{-1}\left(-\frac{1}{2}\right)\right]$ is equal to:
(A) $\frac{1}{2}$
(B) $\frac{1}{3}$
(C) $\frac{1}{4}$
(D) 1

Answer.

$\sin ^{-1}\left(-\frac{1}{2}\right)=\sin ^{-1}\left(-\sin \frac{\pi}{6}\right)=\sin ^{-1}\left[\sin \left(-\frac{\pi}{6}\right)\right]$
$
\begin{aligned}
& =-\frac{\pi}{6} \\
& \therefore \sin \left[\frac{\pi}{3}-\sin ^{-1}\left(-\frac{1}{2}\right)\right]
\end{aligned}
$

$
\begin{aligned}
& =\sin \left[\frac{\pi}{3}-\left(-\frac{\pi}{6}\right)\right] \\
& =\sin \left[\frac{\pi}{3}+\frac{\pi}{6}\right] \\
& =\sin \frac{3 \pi}{6}=\sin \frac{\pi}{2}=1
\end{aligned}
$

Therefore, option (D) is correct.

Ex 2.2 Question 15:

$\tan ^{-1} \sqrt{3}-\cot ^{-1}(-\sqrt{3})$ is equal to:
(A) $\pi$
(B) $-\frac{\pi}{2}$
(C) 0
(D) $2 \sqrt{3}$

Answer.

$\tan ^{-1} \sqrt{3}-\cot ^{-1}(-\sqrt{3})$
$
\begin{aligned}
& =\tan ^{-1} \tan \frac{\pi}{3}-\cot ^{-1}\left(-\cot \frac{\pi}{6}\right) \\
& =\frac{\pi}{3}-\cot ^{-1}\left[\cot \left(\pi-\frac{\pi}{6}\right)\right] \\
& =\frac{\pi}{3}-\left(\pi-\frac{\pi}{6}\right) \\
& =\frac{\pi}{3}-\frac{5 \pi}{6} \\
& =\frac{2 \pi-5 \pi}{6} \\
& =-\frac{3 \pi}{6}=-\frac{\pi}{2}
\end{aligned}
$

Therefore, option (B) is correct.