Find the value of frac{d}{dx} [sin^2 { cot^{- | vedclass

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(B) Let $y = \sin^2 \{ \cot^{-1} \sqrt{\frac{1-x}{1+x}} \}$.Let $x = \cos 	heta$, then $	heta = \cos^{-1} x$.$\sqrt{\frac{1-\cos 	heta}{1+\cos 	he

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$\frac{1}{2}$

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(B) Let $y = \sin^2 \{ \cot^{-1} \sqrt{\frac{1-x}{1+x}} \}$.Let $x = \cos 	heta$, then $	heta = \cos^{-1} x$.$\sqrt{\frac{1-\cos 	heta}{1+\cos 	heta}} = \sqrt{\frac{2\sin^2(	heta/2)}{2\cos^2(	heta/2)}} = 	an(	heta/2)$.So, $\cot^{-1} \sqrt{\frac{1-x}{1+x}} = \cot^{-1}(	an(	heta/2)) = \cot^{-1}(\cot(\frac{\pi}{2} - \frac{	heta}{2})) = \frac{\pi}{2} - \frac{	heta}{2}$.Substituting this back, $y = \sin^2(\frac{\pi}{2} - \frac{	heta}{2}) = \cos^2(	heta/2) = \frac{1+\cos 	heta}{2}$.Since $x = \cos 	heta$, we have $y = \frac{1+x}{2} = \frac{1}{2} + \frac{x}{2}$.Now, $\frac{dy}{dx} = \frac{d}{dx}(\frac{1}{2} + \frac{x}{2}) = 0 + \frac{1}{2} = \frac{1}{2}$.

Find the value of $\frac{d}{dx} [\sin^2 \{ \cot^{-1} \sqrt{\frac{1-x}{1+x}} \}]$.

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