if

Let $x=-\tan y$

where $0

Then,

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

$=\sin ^{-1}\{-\sin (2 y)\}+\cos ^{-1}\{\cos (2 y)\}$

$=-\sin ^{-1}\{\sin (2 y)\}+\cos ^{-1}\{\cos (2 y)\}$

$=-2 y+2 y$

$=0$

$\therefore \sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right)+\cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)=0$