Solution

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

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

We have,

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

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

$=3 \theta$

$=3 \sin ^{-1} x$

$=$ L.H.S.

Solution

To prove:

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

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

We have,

$ \begin{aligned} \text{ R.H.S. } & =\cos ^{-1}(4 x^{3}-3 x) \\ & =\cos ^{-1}(4 \cos ^{3} \theta-3 \cos \theta) \\ & =\cos ^{-1}(\cos 3 \theta) \\ & =3 \theta \\ & =3 \cos ^{-1} x \\ & =\text{ L.H.S. } \end{aligned} $

Solution

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

Solution

$\tan ^{-1}(\sqrt{\frac{1-\cos x}{1+\cos x}}), x<\pi$

$\tan ^{-1}(\sqrt{\frac{1-\cos x}{1+\cos x}})=\tan ^{-1}(\sqrt{\frac{2 \sin ^{2} \frac{x}{2}}{2 \cos ^{2} \frac{x}{2}}})$

$=\tan ^{-1}(\frac{\sin \frac{x}{2}}{\cos \frac{x}{2}})=\tan ^{-1}(\tan \frac{x}{2})$

$=\frac{x}{2}$

Solution

$ \begin{aligned} & \tan ^{-1}(\frac{\cos x-\sin x}{\cos x+\sin x}) \\ & =\tan ^{-1}(\frac{1-\frac{\sin x}{\cos x}}{1+\frac{\sin x}{\cos x}}) \\ & =\tan ^{-1}(\frac{1-\tan x}{1+\tan x}) \quad[\tan ^{-1} \frac{x-y}{1-x y}=\tan ^{-1} x-\tan ^{-1} y] \\ & =\tan ^{-1}(1)-\tan ^{-1}(\tan x) \quad \\ & =\frac{\pi}{4}-x \end{aligned} $

Solution

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

Put $x=a \sin \theta \Rightarrow \frac{x}{a}=\sin \theta \Rightarrow \theta=\sin ^{-1}(\frac{x}{a})$

$\therefore \tan ^{-1} \frac{x}{\sqrt{a^{2}-x^{2}}}=\tan ^{-1}(\frac{a \sin \theta}{\sqrt{a^{2}-a^{2} \sin ^{2} \theta}})$

$=\tan ^{-1}(\frac{a \sin \theta}{a \sqrt{1-\sin ^{2} \theta}})=\tan ^{-1}(\frac{a \sin \theta}{a \cos \theta})$

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

Solution

$ \begin{aligned} & \tan ^{-1}(\frac{3 a^{2} x-x^{3}}{a^{3}-3 a x^{2}}) \\ & \text{ Put } x=a \tan \theta \Rightarrow \frac{x}{a}=\tan \theta \Rightarrow \theta=\tan ^{-1} \frac{x}{a} \\ & \tan ^{-1}(\frac{3 a^{2} x-x^{3}}{a^{3}-3 a x^{2}})=\tan ^{-1}(\frac{3 a^{2} \cdot a \tan \theta-a^{3} \tan ^{3} \theta}{a^{3}-3 a \cdot a^{2} \tan ^{2} \theta}) \\ & =\tan ^{-1}(\frac{3 a^{3} \tan \theta-a^{3} \tan ^{3} \theta}{a^{3}-3 a^{3} \tan ^{2} \theta}) \\ & =\tan ^{-1}(\frac{3 \tan \theta-\tan ^{3} \theta}{1-3 \tan ^{2} \theta}) \\ & =\tan ^{-1}(\tan 3 \theta) \\ & =3 \theta \\ & =3 \tan ^{-1} \frac{x}{a} \end{aligned} $

Solution

Let $\sin ^{-1} \frac{1}{2}=x$. Then, $\sin x=\frac{1}{2}=\sin (\frac{\pi}{6})$.

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

$\therefore \tan ^{-1}[2 \cos (2 \sin ^{-1} \frac{1}{2})]=\tan ^{-1}[2 \cos (2 \times \frac{\pi}{6})]$

$=\tan ^{-1}[2 \cos \frac{\pi}{3}]=\tan ^{-1}[2 \times \frac{1}{2}]$

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

Solution

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

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

Let $y=\tan \Phi$. Then, $\Phi=\tan ^{-1} y$.

$ \begin{aligned} & \therefore \cos ^{-1} \frac{1-y^{2}}{1+y^{2}}=\cos ^{-1}(\frac{1-\tan ^{2} \phi}{1+\tan ^{2} \phi})=\cos ^{-1}(\cos 2 \phi)=2 \phi=2 \tan ^{-1} y \\ & \therefore \tan \frac{1}{2}[\sin ^{-1} \frac{2 x}{1+x^{2}}+\cos ^{-1} \frac{1-y^{2}}{1+y^{2}}] \\ & =\tan \frac{1}{2}[2 \tan ^{-1} x+2 \tan ^{-1} y] \\ & =\tan [\tan ^{-1} x+\tan ^{-1} y] \\ & =\tan [\tan ^{-1}(\frac{x+y}{1-x y})] \\ & =\frac{x+y}{1-x y} \end{aligned} $

Solution

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

We know that $\sin ^{-1}(\sin x)=x$ if $x \in[-\frac{\pi}{2}, \frac{\pi}{2}]$, which is the principal value branch of $\sin ^{-1} x$.

Here, $\frac{2 \pi}{3} \notin[\frac{-\pi}{2}, \frac{\pi}{2}]$

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

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

$\therefore \sin ^{-1}(\sin \frac{2 \pi}{3})=\sin ^{-1}(\sin \frac{\pi}{3})=\frac{\pi}{3}$

Solution

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

We know that $\tan ^{-1}(\tan x)=x$ if $x \in(-\frac{\pi}{2}, \frac{\pi}{2})$, which is the principal value branch of $\tan ^{-1} x$.

Here, $\frac{3 \pi}{4} \notin(\frac{-\pi}{2}, \frac{\pi}{2})$.

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

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

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

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

Solution

Let $\sin ^{-1} \frac{3}{5}=x$. Then, $\sin x=\frac{3}{5} \Rightarrow \cos x=\sqrt{1-\sin ^{2} x}=\frac{4}{5} \Rightarrow \sec x=\frac{5}{4}$.

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

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

$\therefore \sin ^{-1} \frac{3}{5}=\tan ^{-1} \frac{3}{4}$

Now, $\cot ^{-1} \frac{3}{2}=\tan ^{-1} \frac{2}{3}$

…(ii) $\quad[\tan ^{-1} \frac{1}{x}=\cot ^{-1} x]$

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

$=\tan (\tan ^{-1} \frac{3}{4}+\tan ^{-1} \frac{2}{3})$

[Using (i) and (ii)]

$=\tan (\tan ^{-1} \frac{\frac{3}{4}+\frac{2}{3}}{1-\frac{3}{4} \cdot \frac{2}{3}})$

$[\tan ^{-1} x+\tan ^{-1} y=\tan ^{-1} \frac{x+y}{1-x y}]$

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

$=\tan (\tan ^{-1} \frac{17}{6})=\frac{17}{6}$

Solution

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

Here, $\frac{7 \pi}{6} \notin x \in[0, \pi]$.

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

$\cos ^{-1}(\cos \frac{7 \pi}{6})=\cos ^{-1}(\cos \frac{-7 \pi}{6})=\cos ^{-1}[\cos (2 \pi-\frac{7 \pi}{6})] \quad[\cos (2 \pi+x)=\cos x]$

$=\cos ^{-1}[\cos \frac{5 \pi}{6}]$ where $\frac{5 \pi}{6} \in[0, \pi]$

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

The correct answer is $B$.

Solution

Let $\sin ^{-1}(\frac{-1}{2})=x$. Then, $\sin x=\frac{-1}{2}=-\sin \frac{\pi}{6}=\sin (\frac{-\pi}{6})$.

We know that the range of the principal value branch of

$ \sin ^{-1} \text{ is }[\frac{-\pi}{2}, \frac{\pi}{2}] \text{. } $

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

The correct answer is D.

Solution

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

The correct answer is B.