If y = sin^2 left( cot^{-1} left( sqrt{frac{1 | vedclass

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(C) Let $x = \cos 	heta$. Then $	heta = \cos^{-1} x$. Substituting $x = \cos 	heta$ in the expression,we get $\sqrt{\frac{1-\cos 	heta}{1+\cos 	h

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$\frac{1}{2}$

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(C) Let $x = \cos 	heta$. Then $	heta = \cos^{-1} x$. Substituting $x = \cos 	heta$ in the expression,we get $\sqrt{\frac{1-\cos 	heta}{1+\cos 	heta}} = \sqrt{\frac{2\sin^2(	heta/2)}{2\cos^2(	heta/2)}} = 	an(	heta/2)$. So,the expression becomes $y = \sin^2 \left( \cot^{-1} (	an(	heta/2)) ight)$. Since $	an(	heta/2) = \cot(\frac{\pi}{2} - \frac{	heta}{2})$,we have $y = \sin^2 \left( \cot^{-1} \left( \cot \left( \frac{\pi}{2} - \frac{	heta}{2} ight) ight) ight)$. This simplifies to $y = \sin^2 \left( \frac{\pi}{2} - \frac{	heta}{2} ight) = \cos^2(	heta/2)$. Using the identity $\cos^2(	heta/2) = \frac{1+\cos 	heta}{2}$,we get $y = \frac{1+x}{2} = \frac{1}{2} + \frac{x}{2}$. Therefore,$\frac{dy}{dx} = \frac{d}{dx} \left( \frac{1}{2} + \frac{x}{2} ight) = 0 + \frac{1}{2} = \frac{1}{2}$.

If $y = \sin^2 \left( \cot^{-1} \left( \sqrt{\frac{1-x}{1+x}} \right) \right)$,then $\frac{dy}{dx} = $

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