Miscellaneous Exercise (Revised) - Chapter 2 - Inverse Trigonometry - Ncert Solutions class 12 - Maths

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NCERT Class 12 Maths Solutions: Chapter 2 - Inverse Trigonometry

Miscellaneous Exercise Question 1:

Find the value of the following:
$\cos ^{-1}\left(\cos \frac{13 \pi}{6}\right)$

Answer.

$\cos ^{-1}\left(\cos \frac{13 \pi}{6}\right)$
$=\cos ^{-1}\left(\cos \frac{12 \pi+\pi}{6}\right)$
$=\cos ^{-1}\left[\cos \left(2 \pi+\frac{\pi}{6}\right)\right]$
$=\cos ^{-1}\left(\cos \frac{\pi}{6}\right)$
$=\cos ^{-1}\left(\cos \frac{\pi}{6}\right)=\frac{\pi}{6}$
Miscellaneous Exercise Question 2.

$\tan ^{-1}\left(\tan \frac{7 \pi}{6}\right)$

Answer.

$\tan ^{-1}\left(\tan \frac{7 \pi}{6}\right)$
$=\tan ^{-1}\left(\tan \frac{6 \pi+\pi}{6}\right)$

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

Miscellaneous Exercise Question 3.

Prove that: $2 \sin ^{-1} \frac{3}{5}=\tan ^{-1} \frac{24}{7}$

Answer.

Let $\sin ^{-1} \frac{3}{5}=\theta$ so that $\sin \theta=\frac{3}{5}$
$
\begin{aligned}
& \therefore \cos \theta=\sqrt{1-\sin ^2 \theta}=\sqrt{1-\frac{9}{25}}=\sqrt{\frac{16}{25}}=\frac{4}{5} \\
& \therefore \tan \theta=\frac{\sin \theta}{\cos \theta}=\frac{3}{4}
\end{aligned}
$

Since $\tan 2 \theta=\frac{2 \tan \theta}{1-\tan ^2 \theta}$
$
\begin{aligned}
& =\frac{2 \times \frac{3}{4}}{1-\frac{9}{16}}=\frac{\frac{3}{2}}{\frac{7}{16}}=\frac{24}{7} \\
& \Rightarrow 2 \theta=\tan ^{-1} \frac{24}{7} \\
& \Rightarrow 2 \sin ^{-1} \frac{3}{5}=\tan ^{-1} \frac{24}{7}
\end{aligned}
$

Miscellaneous Exercise Question 4.

Prove that: $\sin ^{-1} \frac{8}{17}+\sin ^{-1} \frac{3}{5}=\tan ^{-1} \frac{77}{36}$

Answer.

Let $\sin ^{-1} \frac{8}{17}=\theta$ so that $\sin \theta=\frac{8}{17}$
$
\begin{aligned}
& \therefore \cos \theta=\sqrt{1-\sin ^2 \theta}=\sqrt{1-\frac{64}{289}}=\sqrt{\frac{225}{289}}=\frac{15}{17} \\
& \therefore \tan \theta=\frac{\sin \theta}{\cos \theta}=\frac{8}{15}
\end{aligned}
$

Again, Let $\sin ^{-1} \frac{3}{5}=\phi$ so that $\sin \phi=\frac{3}{5}$
$
\begin{aligned}
& \therefore \cos \phi=\sqrt{1-\sin ^2 \phi}=\sqrt{1-\frac{9}{25}}=\sqrt{\frac{16}{25}}=\frac{4}{5} \\
& \therefore \tan \phi=\frac{\sin \phi}{\cos \phi}=\frac{3}{4}
\end{aligned}
$

Since $\tan (\theta+\phi)=\frac{\tan \theta+\tan \phi}{1-\tan \theta \tan \phi}$
$
\begin{aligned}
& =\frac{\frac{8}{15}+\frac{3}{4}}{1-\frac{8}{15} \times \frac{3}{4}} \\
& =\frac{32+45}{60-24}=\frac{77}{36}
\end{aligned}
$

$\begin{aligned}
& \Rightarrow \theta+\phi=\tan ^{-1} \frac{77}{36} \\
& \Rightarrow \sin ^{-1} \frac{8}{17}+\sin ^{-1} \frac{3}{5}=\tan ^{-1} \frac{77}{36}
\end{aligned}$

Miscellaneous Exercise Question 5.

Prove that: $\cos ^{-1} \frac{4}{5}+\cos ^{-1} \frac{12}{13}=\cos ^{-1} \frac{33}{65}$

Answer.

Let $\cos ^{-1} \frac{4}{5}=\theta$ so that $\cos \theta=\frac{4}{5}$
$
\therefore \sin \theta=\sqrt{1-\cos ^2 \theta}=\sqrt{1-\frac{16}{25}}=\sqrt{\frac{9}{25}}=\frac{3}{5}
$

Again, Let $\cos ^{-1} \frac{12}{13}=\phi$ so that $\cos \phi=\frac{12}{13}$
$
\therefore \sin \phi=\sqrt{1-\cos ^2 \phi}=\sqrt{1-\frac{144}{169}}=\sqrt{\frac{25}{169}}=\frac{5}{13}
$

Since $\cos (\theta+\phi)=\cos \theta \cos \phi-\sin \theta \sin \phi=\frac{4}{5} \times \frac{12}{13}-\frac{3}{5} \times \frac{5}{13}$
$
\begin{aligned}
& =\frac{48-15}{65}=\frac{33}{65} \\
& \Rightarrow \theta+\phi=\cos ^{-1} \frac{33}{65} \\
& \Rightarrow \cos ^{-1} \frac{4}{5}+\cos ^{-1} \frac{12}{13}=\cos ^{-1} \frac{33}{65}
\end{aligned}
$

Miscellaneous Exercise Question 6.

Prove that: $\cos ^{-1} \frac{12}{13}+\sin ^{-1} \frac{3}{5}=\sin ^{-1} \frac{56}{65}$

Answer.

Let $\cos ^{-1} \frac{12}{13}=\theta$ so that $\cos \theta=\frac{12}{13}$
$
\therefore \sin \theta=\sqrt{1-\cos ^2 \theta}=\sqrt{1-\frac{144}{169}}=\sqrt{\frac{25}{169}}=\frac{5}{13}
$

Again, Let $\sin ^{-1} \frac{3}{5}=\phi$ so that $\sin \phi=\frac{3}{5}$

$\begin{aligned}
& \therefore \cos \phi=\sqrt{1-\sin ^2 \phi}=\sqrt{1-\frac{9}{25}}=\sqrt{\frac{16}{25}}=\frac{4}{5} \\
& \text { Since } \sin (\theta+\phi)=\sin \theta \cos \phi+\cos \theta \sin \phi=\frac{5}{13} \times \frac{4}{5}+\frac{12}{13} \times \frac{3}{5} \\
& =\frac{20+36}{65}=\frac{56}{65} \\
& \Rightarrow \theta+\phi=\sin ^{-1} \frac{56}{65} \\
& \Rightarrow \cos ^{-1} \frac{12}{13}+\sin ^{-1} \frac{3}{5}=\sin ^{-1} \frac{56}{65}
\end{aligned}$

Miscellaneous Exercise Question 7.

Prove that: $\tan ^{-1} \frac{63}{16}=\sin ^{-1} \frac{5}{13}+\cos ^{-1} \frac{3}{5}$

Answer.

Let $\sin ^{-1} \frac{5}{13}=\theta$ so that $\sin \theta=\frac{5}{13}$
$
\begin{aligned}
& \therefore \cos \theta=\sqrt{1-\sin ^2 \theta}=\sqrt{1-\frac{25}{169}}=\sqrt{\frac{144}{169}}=\frac{12}{13} \\
& \therefore \tan \theta=\frac{\sin \theta}{\cos \theta}=\frac{5}{12}
\end{aligned}
$

Again, Let $\cos ^{-1} \frac{3}{5}=\phi$ so that $\cos \phi=\frac{3}{5}$
$
\begin{aligned}
& \therefore \sin \phi=\sqrt{1-\cos ^2 \phi}=\sqrt{1-\frac{9}{25}}=\sqrt{\frac{16}{25}}=\frac{4}{5} \\
& \therefore \tan \phi=\frac{\sin \phi}{\cos \phi}=\frac{4}{3}
\end{aligned}
$

$\begin{aligned}
& \text { Since } \tan (\theta+\phi)=\frac{\tan \theta+\tan \phi}{1-\tan \theta \tan \phi}=\frac{\frac{5}{12}+\frac{4}{3}}{1-\frac{5}{12} \times \frac{4}{3}} \\
& =\frac{\frac{12}{4}}{\frac{4}{9}}=\frac{63}{16} \\
& \Rightarrow \theta+\phi=\tan ^{-1} \frac{63}{16} \\
& \Rightarrow \tan ^{-1} \frac{63}{16}=\sin ^{-1} \frac{5}{13}+\cos ^{-1} \frac{3}{5}
\end{aligned}$

Miscellaneous Exercise Question 9.

Prove that: $\tan ^{-1} \sqrt{x}=\frac{1}{2} \cos ^{-1}\left(\frac{1-x}{1+x}\right), x \in[0,1]$

Answer.

Let $\tan ^{-1} \sqrt{x}=\theta$ so that $\tan \theta=\sqrt{x}$
$
\begin{aligned}
& \Rightarrow x=\tan ^2 \theta \\
& \therefore \frac{1}{2} \cos ^{-1}\left(\frac{1-x}{1+x}\right) \\
& =\frac{1}{2} \cos ^{-1}\left(\frac{1-\tan ^2 \theta}{1+\tan ^2 \theta}\right) \\
& =\frac{1}{2} \cos ^{-1} \cos 2 \theta \\
& =\frac{1}{2} \times 2 \theta=\theta \\
& =\tan ^{-1} \sqrt{x}
\end{aligned}
$
Miscellaneous Exercise Question 10.

Prove that: $\cot ^{-1}\left(\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right)=\frac{x}{2}, x \in\left(0, \frac{\pi}{4}\right)$

Answer.

We know that $1+\sin x=\cos ^2 \frac{x}{2}+\sin ^2 \frac{x}{2}+2 \cos \frac{x}{2} \sin \frac{x}{2}=\left(\cos \frac{x}{2}+\sin \frac{x}{2}\right)^2$
Again, $1-\sin x=\cos ^2 \frac{x}{2}+\sin ^2 \frac{x}{2}-2 \cos \frac{x}{2} \sin \frac{x}{2}$
$
\begin{aligned}
& =\left(\cos \frac{x}{2}-\sin \frac{x}{2}\right)^2 \\
& \therefore \text { L.H.S. }=\cot ^{-1}\left(\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right) \\
& =\cot ^{-1}\left[\frac{\left(\cos \frac{x}{2}+\sin \frac{x}{2}\right)+\left(\cos \frac{x}{2}-\sin \frac{x}{2}\right)}{\left(\cos \frac{x}{2}+\sin \frac{x}{2}\right)-\left(\cos \frac{x}{2}-\sin \frac{x}{2}\right)}\right]
\end{aligned}
$
$
\begin{aligned}
& =\cot ^{-1}\left(\frac{2 \cos \frac{x}{2}}{2 \sin \frac{x}{2}}\right) \\
& =\cot ^{-1} \cot \frac{x}{2}=\frac{x}{2}
\end{aligned}
$

Miscellaneous Exercise Question 11.

Prove that: $\tan ^{-1}\left(\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\right)=\frac{\pi}{4}-\frac{1}{2} \cos ^{-1} x, \frac{-1}{\sqrt{2}} \leq x \leq 1$

Answer.

Putting $x=\cos 2 \theta$ so that $\theta=\frac{1}{2} \cos ^{-1} x$
$
\text { L.H.S. }=\tan ^{-1}\left(\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\right)
$

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

Dividing every term by $\sqrt{2} \cos \theta$,
$
\begin{aligned}
& =\tan ^{-1}\left(\frac{1-\tan \theta}{1+\tan \theta}\right) \\
& =\tan ^{-1}\left(\frac{\tan \frac{\pi}{4}-\tan \theta}{1+\tan \frac{\pi}{4} \tan \theta}\right) \\
& =\tan ^{-1} \tan \left(\frac{\pi}{4}-\theta\right)=\frac{\pi}{4}-\theta \\
& =\frac{\pi}{4}-\frac{1}{2} \cos ^{-1} x=\text { R.H.S. }
\end{aligned}
$

Miscellaneous Exercise Question 13.

Solve the equation: $2 \tan ^{-1}(\cos x)=\tan ^{-1}(2 \operatorname{cosec} x)$

Answer.

$2 \tan ^{-1}(\cos x)=\tan ^{-1}(2 \operatorname{cosec} x)$
$
\begin{aligned}
& \Rightarrow \tan ^{-1}\left(\frac{2 \cos x}{1-\cos ^2 x}\right)=\tan ^{-1}\left(\frac{2}{\sin x}\right) \\
& \Rightarrow \frac{2 \cos x}{1-\cos ^2 x}=\frac{2}{\sin x} \\
& \Rightarrow \frac{\cos x}{\sin x}=1 \\
& \Rightarrow \cot x=1 \\
& \Rightarrow x=\frac{\pi}{4}
\end{aligned}
$
Miscellaneous Exercise Question 14.

Solve the equation: $\tan ^{-1}\left(\frac{1-x}{1+x}\right)=\frac{1}{2} \tan ^{-1} x,(x>0)$

Answer.

Putting $x=\tan \theta$

$\begin{aligned}
& \tan ^{-1}\left(\frac{1-x}{1+x}\right)=\frac{1}{2} \tan ^{-1} x \\
& \Rightarrow \tan ^{-1}\left(\frac{1-\tan \theta}{1+\tan \theta}\right)=\frac{1}{2} \tan ^{-1} \tan \theta \\
& \Rightarrow \tan ^{-1}\left(\frac{\tan \frac{\pi}{4}-\tan \theta}{\tan \frac{\pi}{4}+\tan \theta}\right)=\frac{1}{2} \theta \\
& \Rightarrow \tan ^{-1} \tan \left(\frac{\pi}{4}-\theta\right)=\frac{\theta}{2} \\
& \Rightarrow \frac{\pi}{4}-\theta=\frac{\theta}{2} \\
& \Rightarrow \theta+\frac{\theta}{2}=\frac{\pi}{4} \\
& \Rightarrow \frac{3 \theta}{2}=\frac{\pi}{4} \\
& \Rightarrow 12 \theta=2 \pi \\
& \Rightarrow \theta=\frac{\pi}{6} \\
& \therefore x=\tan \theta=\tan \frac{\pi}{6}=\frac{1}{\sqrt{3}}
\end{aligned}$

Miscellaneous Exercise Question 15.

$\sin \left(\tan ^{-1} x\right),|x|<1$ is equal to:
(A) $\frac{x}{\sqrt{1-x^2}}$

(B) $\frac{1}{\sqrt{1-x^2}}$
(C) $\frac{1}{\sqrt{1+x^2}}$
(D) $\frac{x}{\sqrt{1+x^2}}$

Answer.

Let $\sin \left(\tan ^{-1} x\right)=\sin \theta$ where $\theta=\tan ^{-1} x$ so that $x=\tan \theta$
$
\Rightarrow \sin \left(\tan ^{-1} x\right)=\frac{1}{\operatorname{cosec} \theta}=\frac{1}{\sqrt{1+\cot ^2 \theta}}
$

Putting $\cot \theta=\frac{1}{\tan \theta}=\frac{1}{x}$
$
\Rightarrow \sin \left(\tan ^{-1} x\right)=\frac{1}{\sqrt{1+\frac{1}{x^2}}}=\frac{x}{\sqrt{x^2+1}}
$

Therefore, option (D) is correct.

Miscellaneous Exercise Question 16.

$\sin ^{-1}(1-x)-2 \sin ^{-1} x=\frac{\pi}{2}$, then $x$ is equal to:
(A) $0, \frac{1}{2}$
(B) $1, \frac{1}{2}$
(C) 0
(D) $\frac{1}{2}$

Answer.

Putting $\sin ^{-1} x=\theta$ so that $x=\sin \theta$
$
\begin{aligned}
& \therefore \sin ^{-1}(1-x)-2 \sin ^{-1} x=\frac{\pi}{2} \\
& \Rightarrow \sin ^{-1}(1-x)-2 \theta=\frac{\pi}{2} \\
& \Rightarrow \sin ^{-1}(1-x)=\frac{\pi}{2}+2 \theta \\
& \Rightarrow 1-x=\sin \left(\frac{\pi}{2}+2 \theta\right) \\
& \Rightarrow 1-x=\cos 2 \theta \\
& \Rightarrow 1-x=1-2 \sin ^2 \theta \\
& \Rightarrow 1-x=1-2 x^2[x=\sin \theta] \\
& \Rightarrow-x=-2 x^2 \\
& \Rightarrow 2 x^2-x=0 \\
& \Rightarrow x(2 x-1)=0 \\
& \Rightarrow x=0 \text { or } 2 x-1=0 \\
& \Rightarrow x=0 \text { or } x=\frac{1}{2}
\end{aligned}
$

But $x=\frac{1}{2}$ does not satisfy the given equation.
Therefore, option (C) is correct.

Miscellaneous Exercise Question 17.

$\text {} \tan ^{-1}\left(\frac{x}{y}\right)-\tan ^{-1}\left(\frac{x-y}{x+y}\right) \text { is equal to: }$

(A) $\frac{\pi}{2}$
(B) $\frac{\pi}{3}$
(C) $\frac{\pi}{4}$
(D) $-\frac{3 \pi}{4}$

Answer.

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

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

Therefore, option (C) is correct.