Question

Solve:
${\tan }^{-1}\left(\frac{2x}{1-x^2}\right)$

Answer 1

$tan-\left(\frac{2x}{1-x^2}\right)$

let $ta{n}^{-1}\left(\frac{2x}{1-x^2}\right)=P$

$tanp=\frac{2x}{1-x^2}=$

$\Rightarrow tan2\left(\frac{p}{2}\right)=\frac{2x}{1-x^2}$

$\Rightarrow \frac{2tan\left(\frac{p}{2}\right)}{1-ta{n}^2\left(\frac{p}{2}\right)}=\frac{2x}{1-x^2}$

$\Rightarrow x=\left(tan\frac{p}{2}\right)$

$\Rightarrow ta{n}^1\left(x\right)=\frac{p}{2}\Rightarrow p=2ta{n}^{-1}x$

$\therefore ta{n}^{-1}\left(\frac{2x}{1-x^2}\right)=P=2ta{n}^{-1}x$