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

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

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$\frac{-1}{2}$

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(A) Given $y=\sin ^2\left(\cot ^{-1} \sqrt{\frac{1+x}{1-x}}ight)$.Let $	heta=\cot ^{-1} \sqrt{\frac{1+x}{1-x}}$,then $\cot 	heta = \sqrt{\frac{1+x}{1-x}}$.Squaring both sides,we get $\cot^2 	heta = \frac{1+x}{1-x}$.Using the identity $1+\cot^2 	heta = \csc^2 	heta$,we have $\csc^2 	heta = 1 + \frac{1+x}{1-x} = \frac{1-x+1+x}{1-x} = \frac{2}{1-x}$.Since $\sin^2 	heta = \frac{1}{\csc^2 	heta}$,we get $\sin^2 	heta = \frac{1-x}{2}$.Substituting this back into the expression for $y$,we get $y = \sin^2 	heta = \frac{1-x}{2}$.Now,differentiating with respect to $x$,we get $\frac{dy}{dx} = \frac{d}{dx} \left(\frac{1}{2} - \frac{x}{2}ight) = 0 - \frac{1}{2} = -\frac{1}{2}$.

If $y=\sin ^2\left(\cot ^{-1} \sqrt{\frac{1+x}{1-x}}\right)$,then $\frac{d y}{d x}$ has the value

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