Question

If the sum of all the solutions of $\tan ^{-1}\left(\frac{2 x}{1-x^2}\right)+\cot ^{-1}\left(\frac{1-x^2}{2 x}\right)=\frac{\pi}{3},-1 < x < 1, x \neq 0$, is $\alpha-\frac{4}{\sqrt{3}}$, then $\alpha$ is equal to ___

Answer 1

Correct Answer: 2

The correct answer is 2.
Case $I:x>0$
${\tan }^{-1}\frac{2x}{1-x^2}+{\tan }^{-1}\frac{2x}{1-x^2}=\frac{\pi }{3}$
$x=2-\sqrt{3}$
Case II : $x<0$
${\tan }^{-1}\frac{2x}{1-x^2}+{\tan }^{-1}\frac{2x}{1-x^2}+\pi =\frac{\pi }{3}$
$x=\frac{-1}{\sqrt{3}}\Rightarrow \alpha =2$