Question

If 3 ${\sin }^{-1}\left(\frac{2x}{1+x^2}\right)-4{\cos }^{-1}\left(\frac{1-x^2}{1+x^2}\right)+2{\tan }^{-1}\left(\frac{2x}{1-x^2}\right)=\frac{\mathrm{\pi }}{3}$ is equal to
(a) $\frac{1}{\sqrt{3}}$

(b) $-\frac{1}{\sqrt{3}}$

(c) $\sqrt{3}$

(d) $-\frac{\sqrt{3}}{4}$

Answer 1

(a) $\frac{1}{\sqrt{3}}$

Let $x=\tan y$
Then,
$$\begin{gathered} 3{\sin }^{-1}\left(\frac{2\tan y}{1+{\tan }^{2}y}\right)-4\left(\frac{1-{\tan }^{2}y}{1+{\tan }^{2}y}\right)+2{\tan }^{-1}\left(\frac{2\tan y}{1-{\tan }^{2}y}\right)=\frac{\mathrm{\pi }}{3} \\ \Rightarrow 3{\sin }^{-1}\left(\sin 2y\right)-4{\cos }^{-1}\left(\cos 2y\right)+2{\tan }^{-1}\left(\tan 2y\right)=\frac{\mathrm{\pi }}{3} \\ \left[\because \sin 2y=\left(\frac{2\tan y}{1+{\tan }^{2}y}\right),\cos 2y=\left(\frac{1-{\tan }^{2}y}{1+{\tan }^{2}y}\right)\mathrm{and}\tan 2y=\left(\frac{2\tan y}{1-{\tan }^{2}y}\right)\right] \\ \Rightarrow 3\times 2y-4\times 2y+2\times 2y=\frac{\mathrm{\pi }}{3} \\ \Rightarrow 6y-8y+4y=\frac{\mathrm{\pi }}{3} \\ \Rightarrow 2y=\frac{\mathrm{\pi }}{3} \\ \Rightarrow y=\frac{\mathrm{\pi }}{6} \\ \Rightarrow {\tan }^{-1}x=\frac{\mathrm{\pi }}{6}\left[\because {\tan }^{-1}x=y\right] \\ \Rightarrow x=\tan \frac{\mathrm{\pi }}{6} \\ \Rightarrow x=\frac{1}{\sqrt{3}} \end{gathered}$$