$\frac{d}{d x}\left(\sin ^2 \cot ^{-1} \frac{1}{\sqrt{\frac{1+x}{1-x}}}\right)=$
Answer & explanation
Correct answer: option 3
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.