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

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(A) The equation is given as,$y = \sin^{2} (\cot^{-1} \sqrt{\frac{1 + x}{1 - x}})$.Substitute $x = \cos 2	heta$,then $\frac{1 + x}{1 - x} = \frac{1 +

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$\frac{- 1}{2}$

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(A) The equation is given as,$y = \sin^{2} (\cot^{-1} \sqrt{\frac{1 + x}{1 - x}})$.Substitute $x = \cos 2	heta$,then $\frac{1 + x}{1 - x} = \frac{1 + \cos 2	heta}{1 - \cos 2	heta} = \frac{2\cos^{2}	heta}{2\sin^{2}	heta} = \cot^{2}	heta$.So,$y = \sin^{2} (\cot^{-1} \sqrt{\cot^{2}	heta}) = \sin^{2} (\cot^{-1} (\cot	heta)) = \sin^{2}	heta$.Using the identity $\sin^{2}	heta = \frac{1 - \cos 2	heta}{2}$,we get $y = \frac{1 - x}{2}$.Differentiating with respect to $x$:$\frac{dy}{dx} = \frac{d}{dx} (\frac{1}{2} - \frac{x}{2}) = 0 - \frac{1}{2} = -\frac{1}{2}$.

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

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