Question

Show that ${\sin }^{-1}\left(\frac{1-x^2}{1+x^2}\right)=\frac{\pi }{2}-2{\tan }^{-1}x.$

Answer 1

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

Put $x=\tan t$

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

$={\sin }^{-1}\left(\cos 2t\right)$

$=\frac{\pi }{2}-{\cos }^{-1}\left(\cos 2t\right)$

$=\frac{\pi }{2}-2t$

Put $t={\tan }^{-1}x$

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