If y = cos^2 [cot^{-1} (sqrt{frac{1-x}{1+x}}) | vedclass

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

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$-\frac{1}{2}$

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(D) Let $x = \cos 	heta$, then $	heta = \cos^{-1} x$.$\sqrt{\frac{1-x}{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)$.Now, $\cot^{-1}(	an(	heta/2)) = \cot^{-1}(\cot(\frac{\pi}{2} - \frac{	heta}{2})) = \frac{\pi}{2} - \frac{	heta}{2}$.Substituting this into the expression for $y$:$y = \cos^2(\frac{\pi}{2} - \frac{	heta}{2}) = \sin^2(	heta/2)$.Using the identity $\sin^2(	heta/2) = \frac{1-\cos 	heta}{2}$:$y = \frac{1-x}{2} = \frac{1}{2} - \frac{x}{2}$.Therefore, $\frac{dy}{dx} = \frac{d}{dx}(\frac{1}{2} - \frac{x}{2}) = -\frac{1}{2}$.

If $y = \cos^2 [\cot^{-1} (\sqrt{\frac{1-x}{1+x}})]$, then $\frac{dy}{dx} = \dots$

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