Question

If $3{\sin }^{-1}\left(\frac{2x}{1+x^2}\right)-4{\cos }^{-1}\left(\frac{1-x^2}{1+x^2}\right)+2{\tan }^{-1}\left(\frac{2x}{1-x^2}\right)=\frac{\pi }{3}$, then find the value of $x$

• A. $\frac{1}{\sqrt{3}}$
• B. $-\frac{1}{\sqrt{3}}$
• C. $\sqrt{3}$
• D. $-\frac{\sqrt{3}}{4}$

Answer 1

The correct option is A $\frac{1}{\sqrt{3}}$

given, $3{\sin }^{-1}\left(\frac{2x}{1+x^2}\right)-4{\cos }^{-1}\left(\frac{1-x^2}{1+x^2}\right)+2{\tan }^{-1}\left(\frac{2x}{1-x^2}\right)=\frac{\pi }{3}$

Let $x=\tan \theta$

$\Rightarrow 3{\sin }^{-1}\left(\frac{2\tan \theta }{1+{\tan }^2\theta }\right)-4{\cos }^{-1}\left(\frac{1-{\tan }^2\theta }{1+{\tan }^2\theta }\right)+2{\tan }^{-1}\left(\frac{2\tan \theta }{1-{\tan }^2\theta }\right)=\frac{\pi }{3}$

We know that,

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

$\Rightarrow 3{\sin }^{-1}\left(\sin 2\theta \right)-4{\cos }^{-1}\left(\cos 2\theta \right)+2{\tan }^{-1}\left(\tan 2\theta \right)=\frac{\pi }{3}$

$\Rightarrow 3\times 2\theta -4\times 2\theta +2\times 2\theta =\frac{\pi }{3}$

$\Rightarrow 6\theta -8\theta +4\theta =\frac{\pi }{3}$

$\Rightarrow 2\theta =\frac{\pi }{3}$

$\Rightarrow \theta =\frac{\pi }{6}$ where $x=\tan \theta$

$\Rightarrow {\tan }^{-1}x=\frac{\pi }{6}$ where $\theta ={\tan }^{-1}x$

$\therefore x=\tan \frac{\pi }{6}=\frac{1}{\sqrt{3}}$