Choose the correct alternative in the following:

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

Put $x=\tan \theta \Rightarrow \theta=\tan ^{-1} x$

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

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

$y=\sin ^{-1}\left(\sin \left(\frac{\pi}{2}-2 \theta\right)\right)$

$y=\frac{\pi}{2}-2 \theta$

Putting value of $\theta$ we get,

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

Differentiating w.r.t $\mathrm{x}$ we get,

$\frac{\mathrm{dy}}{\mathrm{dx}}=0-2\left(\frac{1}{1+\mathrm{x}^{2}}\right)$

$\because \frac{\mathrm{d}}{\mathrm{dx}} \tan ^{-1} \mathrm{x}=\frac{1}{1+\mathrm{x}^{2}}$

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}}=-\frac{2}{1+\mathrm{x}^{2}}=(\mathrm{A})$