Question

Find $x$, if $x\in (0,1)$
$3{\sin }^{-1}\left(\frac{2x}{1+x^2}\right)-4{\cos }^{-1}\left(\frac{1-x^2}{1+x^2}\right)+2{\tan }^{-1}\left(\frac{2x}{1-x^2}\right)=\frac{\pi }{3}$

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

Answer 1

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

Put $x=\tan \theta$
$\Rightarrow {\tan }^{-1}x=\theta \in (0,\frac{\pi }{4})\Rightarrow 2\theta \in (0,\frac{\pi }{2})$
Now,
$3{\sin }^{-1}\left(\frac{2\tan \theta }{1+{\tan }^2\theta }\right)-4{\cos }^{-1}\left(\frac{1-{\tan }^2\theta }{1+{\tan }^2\theta }\right)+2{\tan }^{-1}\left(\frac{2\tan \theta }{1-{\tan }^2\theta }\right)=\frac{\pi }{3}\Rightarrow 3{\sin }^{-1}\left(\sin 2\theta \right)-4{\cos }^{-1}\left(\cos 2\theta \right)+2{\tan }^{-1}\left(\tan 2\theta \right)=\frac{\pi }{3}$
Since we know that $2\theta \in (0,\frac{\pi }{2})$, the above expression can be written as
$\Rightarrow 3\cdot 2\theta -4\cdot 2\theta +2\cdot 2\theta =\frac{\pi }{3}\Rightarrow 2\theta =\frac{\pi }{3}\Rightarrow \theta =\frac{\pi }{6}\therefore x=\tan \frac{\pi }{6}=\frac{1}{\sqrt{3}}$