Question

Find the value of x, if ${\tan }^{-1}\left(\frac{2x}{1-x^2}\right)+{\cot }^{-1}\left(\frac{1-x^2}{2x}\right)=\frac{2\pi }{3},x>0$.

Answer 1

${\tan }^{-1}\left(\frac{2x}{1-x^2}\right)+{\cot }^{-1}\left(\frac{1-x^2}{2x}\right)=\frac{2\pi }{3}$

$\Rightarrow {\tan }^{-1}\left(\frac{2x}{1-x^2}\right)+{\tan }^{-1}\left(\frac{2x}{1-x^2}\right)=\frac{2\pi }{3}$

$\Rightarrow 2{\tan }^{-1}\left(\frac{2x}{1-x^2}\right)=\frac{2\pi }{3}$

$\Rightarrow {\tan }^{-1}\left(\frac{2x}{1-x^2}\right)=\frac{\pi }{3}$

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

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

$\Rightarrow 2x=\sqrt{3}-\sqrt{3}{x}^2$

$\Rightarrow \sqrt{3}{x}^2+2x-\sqrt{3}=0$

$\Rightarrow x=\frac{-2\pm \sqrt{3+4\times \sqrt{3}\times \sqrt{3}}}{2\sqrt{3}}$

$\Rightarrow x=\frac{-2\pm \sqrt{3+12}}{2\sqrt{3}}$

$\Rightarrow x=\frac{-2\pm \sqrt{15}}{2\sqrt{3}}$

Since $x>0,$ we have $x=\frac{-2+\sqrt{15}}{2\sqrt{3}}$

$\therefore x=\frac{-2+\sqrt{15}}{2\sqrt{3}}$