Question

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

Answer 1

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

Let $x=\tan \theta$

$={\tan }^{-1}\left(\frac{1-{\tan }^2\theta }{1+{\tan }^2\theta }\right)$

$={\tan }^{-1}\left(\frac{{\cos }^2\theta -{\sin }^2\theta }{{\cos }^2\theta +{\sin }^2\theta }\right)$

$={\tan }^{-1}\left(\cos 2\theta \right)$