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 $x$ is equal to.

• A. $\frac{1}{\sqrt{3}}$
• B. $-\frac{1}{\sqrt{3}}$
• C. $\sqrt{3}$
• D. $-\frac{\sqrt{3}}{2}$

Answer 1

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

$3{\sin }^{-1}\left[\frac{2x}{1+x^2}\right]-4{\cos }^{-1}\left[\frac{1-x^2}{1+x^2}\right]+2{\tan }^{-1}\left[\frac{2x}{1-x^2}\right]=\frac{\pi }{3}$

$2{\tan }^{-1}x={\tan }^{-1}\left[\frac{2x}{1-x^2}\right]={\sin }^{-1}\left[\frac{2x}{1+x^2}\right]={\cos }^{-1}\left[\frac{1-x^2}{1+x^2}\right]$

$\therefore$ $3\left[2{\tan }^{-1}x\right]-4\left[2{\tan }^{-1}x\right]+2\left[2{\tan }^{-1}x\right]=\frac{\pi }{3}$

$\Rightarrow 6{\tan }^{-1}x-8{\tan }^{-1}x+4{\tan }^{-1}x=\frac{\pi }{3}$

$\Rightarrow 2{\tan }^{-1}x=\frac{\pi }{3}$

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

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

$\Rightarrow x=\frac{1}{\sqrt{3}}$