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$x=$

• A. $\sqrt{3}$
• B. $\frac{1}{\sqrt{3}}$
• C. $1$
• D. None of these

Answer 1

The correct option is B

$\frac{1}{\sqrt{3}}$

Finding the value of $x$:

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

Put $x=\tan \theta$

$\begin{array}{rcl}3{\sin }^{-1}\left(\frac{2\tan \theta }{1+{\tan }^2\theta }\right)-4{\cos }^{-1}\left(\frac{1-{\tan }^2\theta }{1+{\tan }^2\theta }\right)+2{\tan }^{-1}\left(\frac{2\tan \theta }{1-{\tan }^2\theta }\right)& =& \frac{\pi }{3}\end{array}$

$\Rightarrow$ $\begin{array}{rcl}3{\sin }^{-1}(\sin 2\theta )-4{\cos }^{-1}(\cos 2\theta )+2{\tan }^{-1}(\tan 2\theta )& =& \frac{\pi }{3}\end{array}$

$\Rightarrow$ $\begin{array}{rcl}3\left(2\theta \right)-4\left(2\theta \right)+2\left(2\theta \right)& =& \frac{\pi }{3}\end{array}$

$\Rightarrow$ $\begin{array}{rcl}6\theta -8\theta +4\theta & =& \frac{\pi }{3}\end{array}$

$\Rightarrow$ $\begin{array}{rcl}2\theta & =& \frac{\pi }{3}\end{array}$

$\Rightarrow$ $\begin{array}{rcl}\theta & =& \frac{\pi }{6}\end{array}$

$\Rightarrow$ $\begin{array}{rcl}{\tan }^{-1}x& =& \frac{\pi }{6}\end{array}$

$\begin{array}{rcl}\therefore x& =& \tan \left(\frac{\pi }{6}\right)\\ & =& \frac{\mathbf{1}}{\sqrt{\mathbf{3}}}\end{array}$

Hence, the correct answer is option (B)