If $y=\sin ^2\left(\cot ^{-1} \sqrt{\frac{1+x}{1-x}}\right)$, then $\frac{\mathrm{d} y}{\mathrm{~d} x}$ has the value

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$y=\sin ^2\left(\cot ^{-1} \sqrt{\frac{1+x}{1-x}}\right)$
Let $\theta=\cot ^{-1} \sqrt{\frac{1+x}{1-x}}$ $\begin{array}{ll} \therefore & \cot ^2 \theta=\frac{1+x}{1-x} \\ \therefore & 1+\cot ^2 \theta=\frac{2}{1-x} \\ \therefore & \sin ^2 \theta=\frac{1-x}{2} \\ \therefore & \theta=\sin ^{-1} \sqrt{\frac{1-x}{2}} \\ \therefore & y=\left[\sin \left(\sin ^{-1} \sqrt{\frac{1-x}{2}}\right)\right]^2 \\ \therefore & y=\frac{1-x}{2} \\ \therefore & \frac{\mathrm{~d} y}{\mathrm{~d} x}=\frac{-1}{2} \end{array}$

If $y=\sin ^2\left(\cot ^{-1} \sqrt{\frac{1+x}{1-x}}\right)$, then $\frac{\mathrm{d} y}{\mathrm{~d} x}$ has…

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