frac{d}{dx} left[ sin^2 cot^{-1} left( sqrt{f | vedclass

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(B) Let $y = \sin^2 \cot^{-1} \left( \sqrt{\frac{1-x}{1+x}} ight)$.Put $x = \cos 	heta$,so $	heta = \cos^{-1} x$.Then $y = \sin^2 \cot^{-1} \left(

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$\frac{1}{2}$

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(B) Let $y = \sin^2 \cot^{-1} \left( \sqrt{\frac{1-x}{1+x}} ight)$.Put $x = \cos 	heta$,so $	heta = \cos^{-1} x$.Then $y = \sin^2 \cot^{-1} \left( \sqrt{\frac{1-\cos 	heta}{1+\cos 	heta}} ight) = \sin^2 \cot^{-1} \left( 	an \frac{	heta}{2} ight)$.Since $\cot^{-1}(	an \alpha) = \frac{\pi}{2} - \alpha$,we have $y = \sin^2 \left( \frac{\pi}{2} - \frac{	heta}{2} ight) = \cos^2 \frac{	heta}{2}$.Using the identity $\cos^2 \frac{	heta}{2} = \frac{1+\cos 	heta}{2}$,we get $y = \frac{1+x}{2} = \frac{1}{2} + \frac{x}{2}$.Differentiating with respect to $x$,we get $\frac{dy}{dx} = \frac{d}{dx} \left( \frac{1}{2} + \frac{x}{2} ight) = \frac{1}{2}$.

$\frac{d}{dx} \left[ \sin^2 \cot^{-1} \left( \sqrt{\frac{1-x}{1+x}} \right) \right]$ equals

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