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, $\frac{13 \pi}{6} \notin[0, \pi]$.

Now, $\cos ^{-1}(\cos \frac{13 \pi}{6})$ can be written as:

$\cos ^{-1}(\cos \frac{13 \pi}{6})=\cos ^{-1}[\cos (2 \pi+\frac{\pi}{6})]=\cos ^{-1}[\cos (\frac{\pi}{6})]$, where $\frac{\pi}{6} \in[0, \pi]$.

$\therefore \cos ^{-1}(\cos \frac{13 \pi}{6})=\cos ^{-1}[\cos (\frac{\pi}{6})]=\frac{\pi}{6}$

Solution

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, $\frac{7 \pi}{6} \notin(-\frac{\pi}{2}, \frac{\pi}{2})$.

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

$ \begin{aligned} & \tan ^{-1}(\tan \frac{7 \pi}{6})=\tan ^{-1}[\tan (2 \pi-\frac{5 \pi}{6})] \quad[\tan (2 \pi-x)=-\tan x] \\ & =\tan ^{-1}[-\tan (\frac{5 \pi}{6})]=\tan ^{-1}[\tan (-\frac{5 \pi}{6})]=\tan ^{-1}[\tan (\pi-\frac{5 \pi}{6})] \\ & =\tan ^{-1}[\tan (\frac{\pi}{6})], \text{ where } \frac{\pi}{6} \in(-\frac{\pi}{2}, \frac{\pi}{2}) \\ & \therefore \tan ^{-1}(\tan \frac{7 \pi}{6})=\tan ^{-1}(\tan \frac{\pi}{6})=\frac{\pi}{6} \end{aligned} $

Solution

Let $\sin ^{-1} \frac{3}{5}=x$. Then, $\sin x=\frac{3}{5}$.

$\Rightarrow \cos x=\sqrt{1-(\frac{3}{5})^{2}}=\frac{4}{5}$

$\therefore \tan x=\frac{3}{4}$

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

Now, we have:

$ \begin{aligned} \text{ L.H.S. } & =2 \sin ^{-1} \frac{3}{5}=2 \tan ^{-1} \frac{3}{4} \\ & =\tan ^{-1}(\frac{2 \times \frac{3}{4}}{1-(\frac{3}{4})^{2}}) \quad[2 \tan ^{-1} x=\tan ^{-1} \frac{2 x}{1-x^{2}}] \\ & =\tan ^{-1}(\frac{\frac{3}{2}}{\frac{16-9}{16}})=\tan ^{-1}(\frac{3}{2} \times \frac{16}{7}) \\ & =\tan ^{-1} \frac{24}{7}=\text{ R.H.S. } \end{aligned} $

Solution

Let $\sin ^{-1} \frac{8}{17}=x$. Then, $\sin x=\frac{8}{17} \Rightarrow \cos x=\sqrt{1-(\frac{8}{17})^{2}}=\sqrt{\frac{225}{289}}=\frac{15}{17}$.

$\therefore \tan x=\frac{8}{15} \Rightarrow x=\tan ^{-1} \frac{8}{15}$

$\therefore \sin ^{-1} \frac{8}{17}=\tan ^{-1} \frac{8}{15}$

Now, let $\sin ^{-1} \frac{3}{5}=y$. Then, $\sin y=\frac{3}{5} \Rightarrow \cos y=\sqrt{1-(\frac{3}{5})^{2}}=\sqrt{\frac{16}{25}}=\frac{4}{5}$.

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

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

Now, we have:

$ \begin{aligned} & \text{ L.H.S. }=\sin ^{-1} \frac{8}{17}+\sin ^{-1} \frac{3}{5} \\ & =\tan ^{-1} \frac{8}{15}+\tan ^{-1} \frac{3}{4} \quad[\text{ Using (1) and (2)] } \\ & =\tan ^{-1} \frac{\frac{8}{15}+\frac{3}{4}}{1-\frac{8}{15} \times \frac{3}{4}} \\ & =\tan ^{-1}(\frac{32+45}{60-24}) \quad[\tan ^{-1} x+\tan ^{-1} y=\tan ^{-1} \frac{x+y}{1-x y}] \\ & =\tan ^{-1} \frac{77}{36}=\text{ R.H.S. } \end{aligned} $

Solution

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

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

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

Now, let $\cos ^{-1} \frac{12}{13}=y$. Then, $\cos y=\frac{12}{13} \Rightarrow \sin y=\frac{5}{13}$.

$\therefore \tan y=\frac{5}{12} \Rightarrow y=\tan ^{-1} \frac{5}{12}$

$\therefore \cos ^{-1} \frac{12}{13}=\tan ^{-1} \frac{5}{12}$

Let $\cos ^{-1} \frac{33}{65}=z$. Then, $\cos z=\frac{33}{65} \Rightarrow \sin z=\frac{56}{65}$.

$\therefore \tan z=\frac{56}{33} \Rightarrow z=\tan ^{-1} \frac{56}{33}$

$\therefore \cos ^{-1} \frac{33}{65}=\tan ^{-1} \frac{56}{33}$

Now, we will prove that:

L.H.S. $=\cos ^{-1} \frac{4}{5}+\cos ^{-1} \frac{12}{13}$

$ \begin{aligned} & =\tan ^{-1} \frac{3}{4}+\tan ^{-1} \frac{5}{12} \quad[\text{ Using (1) and (2) }] \\ & =\tan ^{-1} \frac{\frac{3}{4}+\frac{5}{12}}{1-\frac{3}{4} \cdot \frac{5}{12}} \quad[\tan ^{-1} x+\tan ^{-1} y=\tan ^{-1} \frac{x+y}{1-x y}] \\ & =\tan ^{-1} \frac{36+20}{48-15} \\ & =\tan ^{-1} \frac{56}{33} \\ & =\tan ^{-1} \frac{56}{33} \\ & =\text{ R.H.S. } \end{aligned} $

Solution

Let $\sin ^{-1} \frac{3}{5}=x$. Then, $\sin x=\frac{3}{5} \Rightarrow \cos x=\sqrt{1-(\frac{3}{5})^{2}}=\sqrt{\frac{16}{25}}=\frac{4}{5}$.

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

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

Now, let $\cos ^{-1} \frac{12}{13}=y$. Then, $\cos y=\frac{12}{13} \Rightarrow \sin y=\frac{5}{13}$.

$\therefore \tan y=\frac{5}{12} \Rightarrow y=\tan ^{-1} \frac{5}{12}$

$\therefore \cos ^{-1} \frac{12}{13}=\tan ^{-1} \frac{5}{12}$

Let $\sin ^{-1} \frac{56}{65}=z$. Then, $\sin z=\frac{56}{65} \Rightarrow \cos z=\frac{33}{65}$.

$\therefore \tan z=\frac{56}{33} \Rightarrow z=\tan ^{-1} \frac{56}{33}$

$\therefore \sin ^{-1} \frac{56}{65}=\tan ^{-1} \frac{56}{33}$

Now, we have:

$ \begin{matrix} \text{ L.H.S. } & =\cos ^{-1} \frac{12}{13}+\sin ^{-1} \frac{3}{5} & \\ & =\tan ^{-1} \frac{5}{12}+\tan ^{-1} \frac{3}{4} & \text{ [Using (1) and (2)] } \\ & =\tan ^{-1} \frac{\frac{5}{12}+\frac{3}{4}}{1-\frac{5}{12} \cdot \frac{3}{4}} & {[\tan ^{-1} x+\tan ^{-1} y=\tan ^{-1} \frac{x+y}{1-x y}]} \\ & =\tan ^{-1} \frac{20+36}{48-15} \\ & =\tan ^{-1} \frac{56}{33} \\ & =\sin ^{-1} \frac{56}{65}=\text{ R.H.S. } & \text{ [Using (3)] } \end{matrix} $

Solution

Let $\sin ^{-1} \frac{5}{13}=x$. Then, $\sin x=\frac{5}{13} \Rightarrow \cos x=\frac{12}{13}$.

$\therefore \tan x=\frac{5}{12} \Rightarrow x=\tan ^{-1} \frac{5}{12}$

$\therefore \sin ^{-1} \frac{5}{13}=\tan ^{-1} \frac{5}{12}$

Let $\cos ^{-1} \frac{3}{5}=y$. Then, $\cos y=\frac{3}{5} \Rightarrow \sin y=\frac{4}{5}$.

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

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

Using (1) and (2), we have

$ \begin{aligned} \text{ R.H.S. } & =\sin ^{-1} \frac{5}{13}+\cos ^{-1} \frac{3}{5} \\ & =\tan ^{-1} \frac{5}{12}+\tan ^{-1} \frac{4}{3} \\ & =\tan ^{-1}(\frac{\frac{5}{12}+\frac{4}{3}}{1-\frac{5}{12} \times \frac{4}{3}}) \quad[\tan ^{-1} x+\tan ^{-1} y=\tan ^{-1} \frac{x+y}{1-x y}] \\ & =\tan ^{-1}(\frac{15+48}{36-20}) \\ & =\tan ^{-1} \frac{63}{16} \\ & =\text{ L.H.S. } \end{aligned} $

Prove that

Solution

Let $x=\tan ^{2} \theta$. Then, $\sqrt{x}=\tan \theta \Rightarrow \theta=\tan ^{-1} \sqrt{x}$.

$\therefore \frac{1-x}{1+x}=\frac{1-\tan ^{2} \theta}{1+\tan ^{2} \theta}=\cos 2 \theta$

Now, we have:

R.H.S. $=\frac{1}{2} \cos ^{-1}(\frac{1-x}{1+x})=\frac{1}{2} \cos ^{-1}(\cos 2 \theta)=\frac{1}{2} \times 2 \theta=\theta=\tan ^{-1} \sqrt{x}=$ L.H.S.

Solution

Consider $\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}$

$=\frac{(\sqrt{1+\sin x}+\sqrt{1-\sin x})^{2}}{(\sqrt{1+\sin x})^{2}-(\sqrt{1-\sin x})^{2}} \quad$ (by rationalizing)

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

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

$=\cot \frac{x}{2}$

$\therefore$ L.H.S $=\cot ^{-1}(\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}})=\cot ^{-1}(\cot \frac{x}{2})=\frac{x}{2}=$ R.H.S.

Solution

Put $x=\cos 2 \theta$ so that $\theta=\frac{1}{2} \cos ^{-1} x$. Then, we have:

L.H.S. $=\tan ^{-1}(\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}})$

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

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

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

$=\tan ^{-1}(\frac{\cos \theta-\sin \theta}{\cos \theta+\sin \theta})=\tan ^{-1}(\frac{1-\tan \theta}{1+\tan \theta})$

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

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

$=\frac{\pi}{4}-\theta=\frac{\pi}{4}-\frac{1}{2} \cos ^{-1} x=$ R.H.S.

Solution

$ \begin{aligned} & 2 \tan ^{-1}(\cos x)=\tan ^{-1}(2 cosec x) \\ & \Rightarrow \tan ^{-1}(\frac{2 \cos x}{1-\cos ^{2} x})=\tan ^{-1}(2 cosec x) \quad[2 \tan ^{-1} x=\tan ^{-1} \frac{2 x}{1-x^{2}}] \\ & \Rightarrow \frac{2 \cos x}{1-\cos ^{2} x}=2 cosec x \\ & \Rightarrow \frac{2 \cos x}{\sin ^{2} x}=\frac{2}{\sin x} \\ & \Rightarrow \cos x=\sin x \\ & \Rightarrow \tan x=1 \\ & \therefore x=\frac{\pi}{4} \end{aligned} $

Solution

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

$ \tan y=x \Rightarrow \sin y=\frac{x}{\sqrt{1+x^{2}}} . $

$ \begin{aligned} & \therefore y=\sin ^{-1}(\frac{x}{\sqrt{1+x^{2}}}) \Rightarrow \tan ^{-1} x=\sin ^{-1}(\frac{x}{\sqrt{1+x^{2}}}) \\ & \therefore \sin (\tan ^{-1} x)=\sin (\sin ^{-1} \frac{x}{\sqrt{1+x^{2}}})=\frac{x}{\sqrt{1+x^{2}}} \end{aligned} $

The correct answer is D.

Solution

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

$\Rightarrow-2 \sin ^{-1} x=\frac{\pi}{2}-\sin ^{-1}(1-x)$

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

Let $\sin ^{-1} x=\theta \Rightarrow \sin \theta=x \Rightarrow \cos \theta=\sqrt{1-x^{2}}$.

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

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

Therefore, from equation (1), we have

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

Put $x=\sin y$. Then, we have:

$-2 \cos ^{-1}(\sqrt{1-\sin ^{2} y})=\cos ^{-1}(1-\sin y)$

$\Rightarrow-2 \cos ^{-1}(\cos y)=\cos ^{-1}(1-\sin y)$

$\Rightarrow-2 y=\cos ^{-1}(1-\sin y)$

$\Rightarrow 1-\sin y=\cos (-2 y)=\cos 2 y$

$\Rightarrow 1-\sin y=1-2 \sin ^{2} y$

$\Rightarrow 2 \sin ^{2} y-\sin y=0$

$\Rightarrow \sin y(2 \sin y-1)=0$

$\Rightarrow \sin y=0$ or $\frac{1}{2}$

$\therefore x=0$ or $x=\frac{1}{2}$

But, when $x=\frac{1}{2}$, it can be observed that:

L.H.S. $=\sin ^{-1}(1-\frac{1}{2})-2 \sin ^{-1} \frac{1}{2}$

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

$ \begin{aligned} & =-\sin ^{-1} \frac{1}{2} \\ & =-\frac{\pi}{6} \neq \frac{\pi}{2} \neq \text{ R.H.S. } \end{aligned} $

$\therefore x=\frac{1}{2}$ is not the solution of the given equation.

Thus, $x=0$.

Hence, the correct answer is $\mathbf{C}$.