The value of frac{d}{d x}left[cos ^{2}left(co | vedclass

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

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$\frac{1}{4}$

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(A) Let $y = \cos ^{2}\left(\cot ^{-1} \sqrt{\frac{2+x}{2-x}}ight)$.Put $x = 2 \cos 	heta$,then $\cos 	heta = \frac{x}{2}$.Then,$\sqrt{\frac{2+x}{2-x}} = \sqrt{\frac{2+2\cos 	heta}{2-2\cos 	heta}} = \sqrt{\frac{2(1+\cos 	heta)}{2(1-\cos 	heta)}} = \sqrt{\frac{2\cos^2(	heta/2)}{2\sin^2(	heta/2)}} = \cot(	heta/2)$.Substituting this into the expression,we get $y = \cos^2(\cot^{-1}(\cot(	heta/2))) = \cos^2(	heta/2)$.Using the identity $\cos^2(	heta/2) = \frac{1+\cos 	heta}{2}$,we have $y = \frac{1}{2} + \frac{1}{2}\cos 	heta$.Since $\cos 	heta = \frac{x}{2}$,we have $y = \frac{1}{2} + \frac{x}{4}$.Now,differentiating with respect to $x$: $\frac{dy}{dx} = \frac{d}{dx}(\frac{1}{2} + \frac{x}{4}) = 0 + \frac{1}{4} = \frac{1}{4}$.

The value of $\frac{d}{d x}\left[\cos ^{2}\left(\cot ^{-1} \sqrt{\frac{2+x}{2-x}}\right)\right]$ is

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