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

$\frac{1}{2}$

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

∴ $=\sin ^2 \cot ^{-1} \frac{1}{\sqrt{\frac{1+\cos \theta}{1-\cos \theta}}} =\sin ^2 \cot ^{-1} \frac{1}{\cot \frac{\theta}{2}}$

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

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

$=\frac{1+\cos \theta}{2}=\frac{1+x}{2}$

$\Rightarrow \frac{d y}{d x}=\frac{1}{2}$

Hence (3) is correct answer.