If `y=sin^(-1)((2x)/(1+x^2))+sec^(-1)((1+x^2)/(1-x^2)),\ \ 0

Explanation

Given `y=\sin ^{-1}(2 x /(1+x^{2})+\sec ^{-1}(1+x^{2}) /(1-x^{2}).`



Put `x=\tan \theta`
$$ \theta=\tan ^{-1} x $$
So `y=\sin ^{-1}[(2 \tan \theta) /(1+\tan ^{2} \theta)]+\cos ^{-1}[(1-\tan ^{2} \theta) /(1+\tan ^{2} \theta)]`
$$
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