Inverse Trigonometric Functions Test - 18

Given,

\(\cot ^{-1}(\sqrt{3})\)

Let \(\cot ^{-1}(\sqrt{3})=\theta\)

\(\Rightarrow \cot \theta=\sqrt{3}=\cot \frac{\pi}{6}\)

\(\therefore \theta=\frac{\pi}{6}\)

Hence, the correct option is (C).

Given,

\(\sec ^{-1}\left[\frac{x^{2}+1}{x^{2}-1}\right]\)

Put \(x=\tan \theta\)

\(=\sec ^{-1}\left[\frac{\tan ^{2} \theta+1}{\tan ^{2} \theta-1}\right]\)

\(=\sec ^{-1}\left[\frac{\frac{\operatorname{ate}^{2} \theta+\operatorname{cas}^{2} \theta}{\cos ^{2} \theta}}{\frac{\sin ^{2} \theta-\cos ^{2} \theta}{\cos ^{2} \theta}}\right]\)

\(=\sec ^{-1}\left[\frac{\sin ^{2} \theta+\cos ^{2} \theta}{\sin ^{2} \theta-\cos ^{2} \theta}\right]\)

\(=\sec ^{-1}\left[\frac{1}{-\left(\cos ^{2} \theta-\sin ^{2} \theta\right)}\right] \quad\left(\because \sin ^{2} \theta+\cos ^{2} \theta=1\right)\)

\(=\sec ^{-1}\left[\frac{1}{-\cos 2 \theta}\right] \quad\left(\because \cos ^{2} \theta-\sin ^{2} \theta=\cos 2 \theta\right)\)

\(=\sec ^{-1}[-\sec 2 \theta] \quad\left(\because \frac{1}{\cos \theta}=\sec \theta\right)\)

\(=\pi-\sec ^{-1}[\sec 2 \theta] \quad\left(\because \sec ^{-1}(-\theta)=\pi-\sec ^{-1} \theta\right)\)

\(=\pi-2 \theta \quad\left(\because \sec ^{-1}(\sec \theta)=\theta\right)\)

Replace \(\theta\) in terms of \(\mathrm{x}_{1}\)

\(x=\tan \theta \Rightarrow \theta=\tan ^{-1} x \)

\(=\pi-2 \tan ^{-1} x\)

\(=\pi-2\left(\frac{ { \pi}}{2}-\cot ^{-1} x\right)\left(\because \tan ^{-1} x+\cot ^{-1} x=\frac{\pi}{2}\right) \)

\(=\pi-\pi+2 \cot ^{-1} x \)

\(=2 \cot ^{-1} x\)

Hence, the correct option is (C).

Given,

\(\sin ^{-1}\left(\frac{2 a}{1+a^{2}}\right)+\sin ^{-1}\left(\frac{2 b}{1+b^{2}}\right)=2 \tan ^{-1} x\)

Let \(\mathrm{a}=\tan \mathrm{y}_{1}\) and \(\mathrm{b}=\tan \mathrm{y}_{2}\)

Therefore, the given equation becomes:

\(\sin ^{-1}\left(\frac{2 a}{1+a^{2}}\right)+\sin ^{-1}\left(\frac{2 b}{1+b^{2}}\right)=2 \tan ^{-1} x\)

\(\sin ^{-1}\left(\frac{2\left(\tan y_{1}\right)}{1+\left(\tan y_{1}\right)^{2}}\right)+\sin ^{-1}\)\(\left(\frac{2\left(\tan y_{2}\right)}{1+\left(\tan y_{2}\right)^{2}}\right)=2 \tan ^{-1} x \)

\(\sin ^{-1}\left(\sin 2 y_{1}\right)+\sin ^{-1}\left(\sin 2 y_{2}\right)=2 \tan ^{-1} x\)

\(2 y_{1}+2 y_{2}=2 \tan ^{-1} x \)

\(y_{1}+y_{2}=\tan ^{-1} x \)

\(\tan \left(y_{1}+y_{2}\right)=x \)

\(\frac{\tan y_{1}+\tan y_{2}}{1-\tan y_{1} \tan y_{2}}=x\)\(\quad\quad(\because \tan (x+y) =\frac{\tan x+\tan y}{1-\tan x \tan y})\)

\(\frac{a+b}{1-a b}=x\)

\(\therefore x=\frac{a+b}{1-a b}\)

Hence, the correct option is (D).

Given,

\(\tan \left(2 \tan ^{-1}(\cos x)\right)\)

As we know, \(2 \tan ^{-1} x =\tan ^{-1} \frac{2 x }{1- x ^{2}}\)

Therefore,

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

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

\(=\tan ^{-1}(2 \cot x \operatorname{cosec} x)\)

\(\tan \left(\tan ^{-1}(2 \cot x \operatorname{cosec} x)\right)=2 \cot x \operatorname{cosec} x \quad\left(\because \tan \left(\tan ^{-1} x\right)=x\right)\)

Hence, the correct option is (B).

Given,

\(\sin ^{-1}\left(\frac{-1}{\sqrt{2}}\right)\)

As we know \(\sin ^{-1}(-x)=-\sin ^{-1} x\)

So, \(\sin ^{-1}\left(\frac{-1}{\sqrt{2}}\right)=-\sin ^{-1}\left(\frac{1}{\sqrt{2}}\right)\)

Let \(\sin ^{-1}\left(\frac{1}{\sqrt{2}}\right)=\theta\)

\(\Rightarrow \sin \theta=\frac{1}{\sqrt{2}}=\sin 45^{\circ}\)

\(\therefore \theta=-45^{\circ}\)

Hence, the correct option is (C).

Given

\(\alpha=\cos ^{-1}\left(\frac{3}{5}\right), \beta=\tan ^{-1}\left(\frac{1}{3}\right)\) \(\Rightarrow \cos \alpha=\frac{3}{5}\) and \(\tan \beta=\frac{1}{3}\)

According to Pythagoras triplet, \(\sin \alpha=\frac{4}{5}\)

Now, \(\tan \alpha=\frac{\sin \alpha}{\cos \alpha}=\frac{\frac{4}{5}}{\frac{3}{5}}=\frac{4}{3}\)

We know that, \(\tan (\alpha-\beta)=\frac{\tan \alpha-\tan \beta}{1+\tan \alpha-\tan \beta}\)

\(\Rightarrow \tan (\alpha-\beta)=\frac{\frac{4}{3}-\frac{1}{3}}{1+\frac{4}{9}}\)

\(\Rightarrow \tan (\alpha-\beta)=\frac{\frac{3}{3}}{\frac{13}{9}} \)

\(\Rightarrow \tan (\alpha-\beta)=\frac{9}{13} \)

\(\Rightarrow \frac{\sin (\alpha-\beta)}{\cos (\alpha-\beta)}=\frac{9}{13}\)

According to options,

\(\frac{\sin (\alpha-\beta)}{\cos (\alpha-\beta)}=\frac{9 / 5 \sqrt{10}}{13 / 5 \sqrt{10}}\)

Now, \(\sin (\alpha-\beta)=\frac{9}{5 \sqrt{10}}\)

\(\therefore \alpha-\beta=\sin ^{-1}\left(\frac{9}{5 \sqrt{10}}\right)\)

In terms of cosine, \(\cos (\alpha-\beta)=\frac{13}{5 \sqrt{10}}\)

\(\therefore \alpha-\beta=\cos ^{-1}\left(\frac{13}{5 \sqrt{10}}\right)\)

In terms of \(\tan , \tan (\alpha-\beta)=\frac{9}{13}\)

\(\therefore \alpha-\beta=\tan ^{-1}\left(\frac{9}{13}\right)\)

Here, \(\alpha-\beta=\sin ^{-1}\left(\frac{9}{5 \sqrt{10}}\right)\)

Hence, the correct option is (D).

Given,

\(\operatorname{cosec}^{2} \cot ^{-1}\left(\frac{5}{12}\right)\)

Let \(\theta=\cot ^{-1}\left(\frac{5}{12}\right)\)

\(\Rightarrow \cot \theta=\frac{5}{12}\)

Therefore,

\(\operatorname{cosec}^{2} \cot ^{-1}\left(\frac{5}{12}\right)=\operatorname{cosec}^{2} \theta\)

\(=1+\cot ^{2} \theta\)\(\quad\quad(\because 1+\cot ^{2} \theta=\operatorname{cosec}^{2} \theta)\)

\(=1+\left(\frac{5}{12}\right)^{2}\)

\(=\frac{169}{144}\)

Hence, the correct option is (B).

We know that \(\sin ^{2} \theta+\cos ^{2} \theta=1\)

\(\Rightarrow 1+\frac{\cos ^{2} \theta}{\sin ^{2} \theta}=\frac{1}{\sin ^{2} \theta}\)

\(\Rightarrow 1+\cot ^{2} \theta=\frac{1}{\sin ^{2} \theta} \)

\(\Rightarrow \sin \theta=\frac{1}{\sqrt{1+\cot ^{2} \theta}}\)

Let \(\cot ^{-1} x=\theta\)

\(\Rightarrow \cot \left(\cot ^{-1} x\right)=\cot \theta \)

\(\Rightarrow \cot \theta=x \)

Therefore,

\(\sin \theta=\frac{1}{\sqrt{1+\cot ^{2} \theta}}\)

\(=\frac{1}{\sqrt{1+x^{2}}} \)

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

Hence, the correct option is (C).

Given,

\(2 \tan ^{-1}\left[\operatorname{cosec}\left(\tan ^{-1} x\right)-\tan \left(\cot ^{1} x\right)\right]\)

\(\Rightarrow 2 \tan ^{-1}\left[\operatorname{cosec}\left(\tan ^{-1} x\right)-\tan \left(90-\tan ^{-1} x\right)\right]\)

Let \(\tan ^{-1} x=\theta\)

\(\Rightarrow 2 \tan ^{-1}[\operatorname{cosec} \theta-\tan (90-\theta)]\)

\(\Rightarrow 2 \tan ^{-1}[\operatorname{cosec} \theta-\cot \theta]\)\(\quad\quad(\because \tan(90 - \theta = \cot \theta)\)

\(\Rightarrow 2 \tan ^{-1}\left[\frac{1}{\sin \theta}-\frac{\cos \theta}{\sin \theta}\right]\)\(\quad\quad(\because \operatorname{cosec} \theta = \frac{1}{\sin \theta}, \cot \theta = \frac{\cos \theta}{\sin \theta}) \)

\(\Rightarrow 2 \tan ^{-1}\left[\frac{1-\cos \theta}{\sin \theta}\right]\)

\(\Rightarrow 2 \tan ^{-1}\left[\frac{2 \sin ^{2} \frac{\theta}{2}}{2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}}\right]\)

\(\Rightarrow 2 \tan ^{-1}\left[\tan \frac{\theta}{2}\right]\)

\(\Rightarrow \theta=\tan ^{-1} x\)

Hence, the correct option is (C).

Given,

\(\cos \left\{\cos ^{-1} \frac{4}{5}+\cos ^{-1} \frac{12}{13}\right\}\)

\(=\cos ^{-1}\left[\frac{4}{5} \times \frac{12}{13}-\sqrt{1-\left(\frac{4}{5}\right)^{2}} \sqrt{1-\left(\frac{12}{13}\right)^{2}}\right] \) \(\quad\quad (\because cos^{-1}x + cos^{-1}y =  cos^{-1}(xy - \sqrt{1−x^{2}} × \sqrt{1−y^{2}}))\)

\(=\cos ^{-1}\left[\frac{48}{65}-\frac{3}{5} \times \frac{5}{13}\right] \)

\(=\cos ^{-1}\left[\frac{48}{65}-\frac{15}{65}\right] \)

\(=\cos ^{-1} \frac{33}{65}\)

Now,

\(\cos \left\{\cos ^{-1} \frac{4}{5}+\cos ^{-1} \frac{12}{13}\right\} \)

\(=\cos \left(\cos ^{-1} \frac{33}{65}\right) \)

\(=\frac{33}{65} \quad\left(\because \cos \left(\cos ^{-1} x\right)=x\right)\)

Hence, the correct option is (B).

Given,

\(\sin ^{-1}(\frac{4}{5})-\sin ^{-1}(\frac{3}{5})\)

\(=\sin ^{-1}\left(\left(\frac{4}{5} \sqrt{1-\left(\frac{3}{5}\right)^{2}}\right)-\left(\frac{3}{5} \sqrt{1-\left(\frac{4}{5}\right)^{2}}\right)\right)\)\(\quad\quad(\because\sin ^{-1} x-\sin ^{-1} y=\sin ^{-1}(x \sqrt{1-y^{2}}-y \sqrt{1-x^{2}})\)

\(=\sin ^{-1}\left(\left(\frac{4}{5} \sqrt{1-\frac{9}{25}}\right)-\left(\frac{3}{5} \sqrt{\left.\left.1-\frac{16}{25}\right)\right)}\right.\right.\)

\(=\sin ^{-1}\left(\left(\frac{4}{5} \sqrt{\frac{25-9}{25}}\right)-\left(\frac{3}{5} \sqrt{\frac{25-16}{25}}\right)\right)\)

\(=\sin ^{-1}\left(\left(\frac{4}{5} \sqrt{\frac{16}{25}}\right)-\left(\frac{3}{5} \sqrt{\frac{9}{25}}\right)\right)\)

\(=\sin ^{-1}\left(\left(\frac{4}{5} \sqrt{\left(\frac{4}{5}\right)^{2}}\right)-\left(\frac{3}{5} \sqrt{\left(\frac{3}{5}\right)^{2}}\right)\right)\)

\(=\sin ^{-1}\left(\left(\frac{4}{5} \times \frac{4}{5}\right)-\left(\frac{3}{5} \times \frac{3}{5}\right)\right)\)

\(=\sin ^{-1}\left(\left(\frac{16}{25}\right)-\left(\frac{9}{25}\right)\right)\)

\(=\sin ^{-1}\left(\frac{7}{25}\right)\)

Hence, the correct option is (B).

Given,

\(\cot \left[\tan ^{-1} \frac{1}{2}+\tan ^{-1} \frac{1}{8}\right]=\)

\(=\cot [\tan ^{-1}\left(\frac{\frac{1}{2}+\frac{1}{8}}{1-\frac{1}{2} \times \frac{1}{8}}\right)] \quad\left(\because \tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{\tan x+\tan y}{1-\tan x \tan y}\right)\right)\)

\(=\cot [\tan ^{-1} \frac{\left(\frac{2+8}{16}\right)}{\left(\frac{16-1}{16}\right)}]\)

\(=\cot [\tan ^{-1} \frac{\left(\frac{10}{16}\right)}{\left(\frac{15}{16}\right)} ]\)

\(=\cot [\tan ^{-1} \frac{10}{15}]\)\(=\tan ^{-1} \frac{2}{3}\)

\(\tan ^{-1} \frac{2}{3}=\cot ^{-1} \frac{3}{2}\)\(\quad\quad(\because \tan ^{-1} x=\cot ^{-1} \frac{1}{x})\) 

Now,

\(\cot \left[\tan ^{-1} \frac{1}{2}+\tan ^{-1} \frac{1}{8}\right] \)

\(=\cot [\cot ^{-1} \frac{3}{2}]\)

\(=\frac{3}{2}\)

Hence, the correct option is (A).

Given

\(\sin ^{-1} \frac{3}{x}+\sin ^{-1} \frac{9}{x}=\frac{\pi}{2}\)

Let

\(\sin ^{-1} \frac{9}{x}=y\)...(i)

\(\Rightarrow \operatorname{siny}=\frac{9}{x}\)....(ii)

Therefore,

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

By using equation (ii) we get,

\(\cos ^{2} y=1-\left(\frac{9}{x}\right)^{2}\)

\(\Rightarrow \cos ^{2} y=1-\frac{81}{x^{2}}\)

\(\Rightarrow \cos ^{2} y=\frac{\sqrt{x^{2}-81}}{x} \)\(\Rightarrow y=\cos ^{-1} \frac{\sqrt{x^{2}-81}}{x}\)

Using equation (ii) we get,

\(\sin ^{-1} \frac{9}{x}=\cos ^{-1} \frac{\sqrt{x^{2}-81}}{x}\)

\(\sin ^{-1} \frac{9}{x}+\cos ^{-1} \frac{\sqrt{x^{2}-81}}{x}=\frac{\pi}{2}\)

Now we know \(\sin ^{-1} \mathrm{y}+\cos ^{-1} \mathrm{y}=\frac{\pi}{2}\)

So,

\(\frac{9}{x}=\frac{\sqrt{x^{2}-81}}{x}\)

\(9=x^{2}-81\)

\(\Rightarrow x^{2}=90\)

\(x=3 \sqrt{10}\)

Hence, the correct option is (B).

Given,

\(\tan ^{-1} 2, \tan ^{-1} 3\) are two angles of a triangle.

We know that

\(\tan ^{-1} \mathrm{a}+\tan ^{-1} \mathrm{~b}=\tan ^{-1}\left(\frac{\mathrm{a}+\mathrm{b}}{1-\mathrm{ab}}\right)\)

Consider

\(\tan ^{-1} 2+\tan ^{-1} 3=\tan ^{-1}\left(\frac{2+3}{1-2.3}\right) \)

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

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

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

Now, the sum of angles of the triangle is \(180^{\circ}\) i.e \(\pi\)

Consider the third angle as \(A\).

\(\tan ^{-1} 2+\tan ^{-1} 3+A=\pi\)

\(\frac{3 \pi}{4}+A=\pi\)

\(\mathrm{A}=\frac{\pi}{4}\)

If \(\tan ^{-1} 2, \tan ^{-1} 3\) are two angles of a triangle, then the third angle is \(\frac{\pi}{4}\)

Hence, the correct option is (C).

Given,

\(\sin ^{-1} \mathrm{x}+\cos ^{-1} \mathrm{y}=\frac{2 \pi}{5}\)

We know that, \(\sin ^{-1} \mathrm{x}+\cos ^{-1} \mathrm{x}=\frac{\pi}{2}\)

\(\sin ^{-1} x=\frac{\pi}{2}-\cos ^{-1} x\)....(i)

Similarly,

\(\sin ^{-1} \mathrm{y}+\cos ^{-1} \mathrm{y}=\frac{\pi}{2}\)

\(\cos ^{-1} \mathrm{y}=\frac{\pi}{2}-\sin ^{-1} \mathrm{y}\)...(i)

Now,

\(\sin ^{-1} \mathrm{x}+\cos ^{-1} \mathrm{y}=\frac{2 \pi}{5}\)

From equation (i) and (ii) we get,

\(\Rightarrow \frac{\pi}{2}-\cos ^{-1} \mathrm{x}+\frac{\pi}{2}-\sin ^{-1} \mathrm{y}=\frac{2 \pi}{5}\)

\(\Rightarrow \pi-\cos ^{-1} \mathrm{x}-\sin ^{-1} \mathrm{y}=\frac{3 \pi}{5}\)

Hence, the correct option is (B).

Given,

\(\tan ^{-1}\left(\frac{8 x \sqrt{x}}{1-16 x^{3}}\right)\)...(i)

Rewrite the equation (i) as follows:

\(\tan ^{-1}\left(\frac{8 x \cdot \sqrt{x}}{1-16 x^{3}}\right)=\tan ^{-1}\left(\frac{2(4 x \sqrt{x})}{1-(4 x \sqrt{x})^{2}}\right)\)

On comparing equation (i) with \(2 \tan ^{-1} \mathrm{x}=\tan ^{-1}(\frac{2 \mathrm{x}}{1-\mathrm{x}^{2}}))\) we get,

\(=2 \tan ^{-1}(4 x \sqrt{x})\)

Hence, the correct option is (A).

Given,

\(\operatorname{cosec} \theta=\sqrt{2 \sqrt{2 \sqrt{2}}} \ldots \ldots \ldots\)

Let \(x=\sqrt{2 \sqrt{2 \sqrt{2}}} \ldots \ldots \ldots\)

On squaring both sides we get,

\(x^{2}=2 \sqrt{2 \sqrt{2}} \ldots \ldots \ldots\)

\(\Rightarrow x^{2}=2 x \quad(\because x=\sqrt{2 \sqrt{2 \sqrt{2}}} \ldots \ldots \ldots .) \)

\(\Rightarrow x^{2}-2 x=0\)

\(\Rightarrow x(x-2)=0 \)

\(\Rightarrow x=2 \text { OR } x=0 \)

\(\text { If } x=2 \text { : } \)

\(\operatorname{cosec} \theta=2\)

\(\Rightarrow \theta=30^{\circ}\)

\(\therefore \tan \theta=\tan 30^{\circ}=\frac{1}{\sqrt{3}}\)

Hence, the correct option is (C).

Given,

\(\tan ^{-1}(1+x)+\tan ^{-1}(1-x)=\frac{\pi}{2}\)

\(\Rightarrow \tan ^{-1}\left(\frac{1+x+1-x}{1-(1+x)(1-x)}\right)=\tan ^{-1}\left(\frac{1}{0}\right) \)\(\quad\quad(\because\tan ^{-1} \mathrm{~A}+\tan ^{-1} \mathrm{~B}=\tan ^{-1}\left(\frac{A+\mathrm{B}}{1-\mathrm{AB}}\right)\) where \(\mathrm{A}>0, \mathrm{~B}>0 \& \mathrm{AB}<1)\)

\(\Rightarrow \frac{2}{1-1+x^{2}}=\frac{1}{0} \)

\(\Rightarrow x^{2}=0 \)

\(\Rightarrow x=0\)

Hence, the correct option is (C).

Let's say that \(\sin ^{-1} 5 x=\theta\)

\(\Rightarrow \sin \left(\sin ^{-1} 5 x\right)=\sin \theta \)

\(\Rightarrow \sin \theta=5 x\)

Since, \(-1 \leq \sin \theta \leq 1\)

\(\Rightarrow-1 \leq 5 x \leq 1\)

\(\Rightarrow-\frac{1}{5} \leq x \leq \frac{1}{5} \)

\(\Rightarrow x \in\left[-\frac{1}{5}, \frac{1}{5}\right]\)

The domain of the function is the closed interval \(\left[-\frac{1}{5}, \frac{1}{5}\right]\).

Hence, the correct option is (C).

Given,

\(\sec (\tan ^{-1} \frac{\mathrm{y}}{2})\)

Let's say that \(\tan ^{-1} \frac{y}{2}=\theta\), where \(\theta \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\).

\(\Rightarrow \tan \theta=\frac{y}{2}\)

\(\Rightarrow \tan ^{2} \theta=\frac{y^{2}}{4}\)

\(\Rightarrow 1+\tan ^{2} \theta=1+\frac{y^{2}}{4}\)

\(=\frac{y^{2}+4}{4}\)

\(\Rightarrow \sec ^{2} \theta=\frac{y^{2}+4}{4}\)\(\quad\quad(\because  1+\tan ^{2} \theta = \sec ^{2} \theta)\)

\(\Rightarrow \sec \theta=\frac{\sqrt{y^{2}+4}}{2}\)

\(\Rightarrow \sec \left(\tan ^{-1} \frac{y}{2}\right)=\frac{\sqrt{y^{2}+4}}{2}\)

Hence, the correct option is (D).

Given,

\(\sin ^{-1} \sin \left(\frac{33 \pi}{5}\right)\)

\(\sin ^{-1} \sin \left(\frac{33 \pi}{5}\right)=\sin ^{-1} \sin \left(6 \pi+\frac{3 \pi}{5}\right) \)

\(=\sin ^{-1} \sin \left(\frac{3 \pi}{5}\right) \quad(: \sin (2 n \pi+\theta)=\sin \theta)\)

\(=\sin ^{-1} \sin \left(\pi-\frac{2 \pi}{5}\right) \)

\(=\sin ^{-1} \sin \left(\frac{2 \pi}{5}\right) \quad(\because \sin (\pi-\theta)=\sin \theta)\)

Here \(\frac{2 \pi}{5}\) is lies between \(\frac{-\pi}{2}\) to \(\frac{\pi}{2}\)

\(\therefore \sin ^{-1} \sin \left(\frac{2 \pi}{5}\right)=\frac{2 \pi}{5}\)

Hence, the correct option is (B).

Given,

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

\(\Rightarrow 2 \tan ^{-1}\left(\frac{1-x}{1+(1 \times x)}\right)=\tan ^{-1} x\)

\(\Rightarrow 2\left[\tan ^{-1} 1-\tan ^{-1} x\right]=\tan ^{-1} x\)

\(\Rightarrow 2\left(\tan ^{-1} 1\right)=3 \tan ^{-1} x\)

\(\Rightarrow 2 \times \frac{\pi}{4}=3 \tan ^{-1} x\)

\(\Rightarrow \frac{\pi}{6}=\tan ^{-1} x\)

\(\Rightarrow x=\tan \frac{\pi}{6}=\frac{1}{\sqrt{3}}\)

Hence, the correct option is (C).

Given,

\(\cos ^{-1}\left(4 x^{3}-3 x\right)\)

Put \(x=\cos \theta\)

\(\Rightarrow \theta = \cos^{-1} x\)

\(\cos ^{-1}\left(4 x^{3}-3 x\right)\)

\(=\cos ^{-1}\left(4 \cos ^{3} \theta-3 \cos \theta\right) \)

\(=\cos ^{-1}(\cos 3 \theta) \quad\left(\because \cos 3 \theta=4 \cos ^{3} \theta-3 \cos \theta\right) \)

\(=3 \theta \quad\left(\because \cos ^{-1} \cos \theta=\theta\right)\)

\(=3 \cos ^{-1} x \)

Hence, the correct option is (A).

Given,

\(x=\tan ^{-1}(\frac{1}{5})\)

\(\tan x=\frac{1}{5}\)

As we know that, \(\sin 2 \theta=\frac{2 \tan \theta}{1+\tan ^{2} \theta}\)

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

\(=\frac{2 \times \frac{1}{3}}{1+\left(\frac{1}{5}\right)^{2}} \)

\(=\frac{\left(\frac{2}{3}\right)}{\frac{s+1}{25}} \)

\(=\frac{2}{5} \times \frac{25}{26}=\frac{10}{26}\)

\(=\frac{5}{13}\)

Hence, the correct option is (B).

Given,

\(\cot ^{-1} \frac{1}{3}-2 \tan ^{-1} \frac{2}{3}\)

\(=\left[\frac{\pi}{2}-\tan ^{-1} \frac{1}{3}\right]-\tan ^{-1} \frac{(\frac{2}{3}+\frac{2}{3})}{(1-\frac{2}{3} \times \frac{2}{3})}\)\(\quad\quad(\because \cot ^{-1} x= \frac{\pi}{2}-\tan ^{-1} x, 2\tan ^{-1} x =\tan ^{-1} (\frac{2x}{1 -x^2}))\)

\(=\frac{\pi}{2}-\left[\tan ^{-1} \frac{12}{5}+\tan ^{-1} \frac{1}{3}\right]\)

\(=\frac{\pi}{2}-\left[\tan ^{-1} \frac{\frac{12}{3}+\frac{1}{1}}{1-\frac{12}{5} \times \frac{1}{3}}\right]\)\(\quad\quad(\because \tan ^{-1} x+\tan ^{-1} y =\tan ^{-1} \frac{x+y}{1 -xy} )\)

\(=\frac{\pi}{2}-\left[\tan ^{-1} \frac{41}{3}\right]\)

\(=\cot ^{-1} \frac{41}{3}\)

Hence, the correct option is (C).

Given,

\(2 \tan ^{-1}(\cos x)=\tan ^{-1}(2 \operatorname{cosec} x)\)

As we know that, \(2 \tan ^{-1} x=\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right),-1 \leq x \leq 1\)

\(2 \tan ^{-1}(\cos x)=\tan ^{-1}\left(\frac{2 \cos x}{1-\cos ^{2} x}\right)\)

As we know that, \(\sin ^{2} x+\cos ^{2} x=1\)

\(\Rightarrow 2 \tan ^{-1}(\cos x)=\tan ^{-1}\left(\frac{2 \cos x}{\sin ^{2} x}\right)\)

\(\Rightarrow \tan ^{-1}\left(\frac{2 \cos x}{\sin ^{2} x}\right)=\tan ^{-1}(2 \operatorname{cosec} x)=\tan ^{-1}(\frac{2}{\sin x})\)

\(\Rightarrow \frac{2 \cos x}{\sin ^{2} x}=\frac{2}{\sin x} \)

\(\Rightarrow \cos x \sin x-\sin ^{2} x=0\)

\(\Rightarrow \sin x(\cos x-\sin x)=0\)

\(\Rightarrow \sin x=0 \text { or } \cos x-\sin x=0 \)

\(\Rightarrow x=0 \text { or } \frac{\pi}{4}\)

As we can see that for \(x=0\) the given equation does not exist Thus, \(x=\frac{\pi }{4}\) is the only solution for the given equation.

Hence, the correct option is (A).

Given,

\(2 \tan ^{-1}\left[2 \cos \left(2 \sin ^{-1} \frac{\sqrt{3}}{2}\right)\right]\)

\(\tan ^{-1}\left[2 \cos \left(2 \sin ^{-1} \frac{\sqrt{3}}{2}\right)\right]\) can be rewritten as:

\(\tan ^{-1}\left[2 \cos \left(2 \sin ^{-1} \frac{\sqrt{3}}{2}\right)\right]=\tan ^{-1}\left[2 \cos \left(\frac{2 \pi}{3}\right)\right]\)

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

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

\(=-\frac{\pi}{4}\)

\(2 \tan ^{-1}\left[2 \cos \left(2 \sin ^{-1} \frac{\sqrt{3}}{2}\right)\right]=-2 \times \frac{\pi}{4}\)

\(=-\frac{\pi}{2}\)

Hence, the correct option is (D).

Given,

\(\sin ^{-1}\left(2 x \sqrt{1-x^{2}}\right)\)

The domain of inverse sin function , \(\sin x\) is \(\in [-1,1]\) 

Domain of the function is calculated as follows:

\(\sin ^{-1}\left(2 x \sqrt{1-x^{2}}\right)\)

\(-1 \leq 2 x \sqrt{1-x^{2}} \leq 1\)

\(-\frac{1}{2} \leq x \sqrt{1-x^{2}} \leq \frac{1}{2}\)

\(x^{2}\left(1-x^{2}\right) \leq \frac{1}{4}\)

Let, \(x^{2} =t\)

\(t\left(1-t\right) \leq \frac{1}{4}\)

\(t-t^{2}-\frac{1}{4} \leq 0\)

\(\left(t-\frac{1}{2}\right)^{2} \leq 0\)

\(t \leq \frac{1}{2}\)

\(x^{2} \leq \frac{1}{{2}}\)

\(x\leq \frac{1}{\sqrt{2}}\)

\(x \in\left[-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right]\)

Hence, the correct option is (C).

Given,

\(\cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)\)

Put \(x=\tan \theta\)

We have to find the value of \(\cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)\)

Put \(x=\tan \theta\)

\(\Rightarrow \cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)=\cos ^{-1}\left(\frac{1-\tan ^{2} \theta}{1+\tan ^{2} \theta}\right)\)

\(\cos ^{-1}\left(\frac{1-\tan ^{2} \theta}{\sec ^{2} \theta}\right) \)\(\quad\quad(\because1+\tan ^{2}\theta=\sec ^{2} \theta)\)

\(=\cos ^{-1}\left(\cos ^{2} \theta-\sin ^{2} \theta\right)\)

\(=\cos ^{-1}(\cos 2 \theta) \quad\left(\because \cos 2 \theta=\cos ^{2} \theta-\sin ^{2} \theta\right)\)

\(=2 \theta \quad\left(\because \cos ^{-1} \cos x=x\right)\)

\(=2 \tan ^{-1} x \quad(\because x=\tan \theta)\)

\(\Rightarrow \cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)=\cos ^{-1}\left(\frac{1-\tan ^{2} \theta}{1+\tan ^{2} \theta}\right) \)

\(\cos ^{-1}\left(\frac{1-\tan ^{2} \theta}{\sec ^{2} \theta}\right)\)

\(=\cos ^{-1}\left(\cos ^{2} \theta-\sin ^{2} \theta\right) \)

\(=\cos ^{-1}(\cos 2 \theta) \quad\left(\because \cos 2 \theta=\cos ^{2} \theta-\sin ^{2} \theta\right) \)

\(=2 \theta \quad\left(\because \cos ^{-1} \cos x=x\right)\)

\(=2 \tan ^{-1} x \quad(\because x=\tan \theta)\)

Hence, the correct option is (C).

Given,

\(\cos \left(2 \sin ^{-1} \sqrt{\frac{1-x}{2}}\right)\)

Substitute \(x=\cos \theta\)

\(\cos \left(2 \sin ^{-1} \sqrt{\frac{1-x}{2}}\right) \)

\(=\cos \left(2 \sin ^{-1} \sqrt{\frac{1-\cos \theta}{2}}\right) \)

\(=\cos \left(2 \sin ^{-1} \sqrt{\left.\frac{2 \sin ^{2}\left(\frac{\theta}{2}\right)}{2}\right)}\right.\)

\(=\cos \left(2 \sin ^{-1} \sin \left(\frac{\theta}{2}\right)\right)\)

\(=\cos \left(2 \times \frac{\theta}{2}\right) \quad\left(\because \sin ^{-1} (\sin x)=x\right)\)

\(=\cos \theta=x\)

Hence, the correct option is (C).