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

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(A) Given $y = \cot^{-1}\left(\sqrt{\frac{1-\sin x}{1+\sin x}}\right)$.Using the identities $1-\sin x = (\cos\frac{x}{2} - \sin\frac{x}{2})^2$ and $1+

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$\frac{1}{2}$

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(A) Given $y = \cot^{-1}\left(\sqrt{\frac{1-\sin x}{1+\sin x}}ight)$.Using the identities $1-\sin x = (\cos\frac{x}{2} - \sin\frac{x}{2})^2$ and $1+\sin x = (\cos\frac{x}{2} + \sin\frac{x}{2})^2$,we get:$y = \cot^{-1}\left(\frac{\cos\frac{x}{2} - \sin\frac{x}{2}}{\cos\frac{x}{2} + \sin\frac{x}{2}}ight)$Dividing numerator and denominator by $\cos\frac{x}{2}$:$y = \cot^{-1}\left(\frac{1 - 	an\frac{x}{2}}{1 + 	an\frac{x}{2}}ight)$Since $\cot^{-1}(z) = 	an^{-1}(\frac{1}{z})$,we have:$y = 	an^{-1}\left(\frac{1 + 	an\frac{x}{2}}{1 - 	an\frac{x}{2}}ight)$Using $	an(\frac{\pi}{4} + 	heta) = \frac{1 + 	an	heta}{1 - 	an	heta}$:$y = 	an^{-1}\left(	an(\frac{\pi}{4} + \frac{x}{2})ight) = \frac{\pi}{4} + \frac{x}{2}$Now,differentiating with respect to $x$:$\frac{dy}{dx} = \frac{d}{dx}(\frac{\pi}{4} + \frac{x}{2}) = 0 + \frac{1}{2} = \frac{1}{2}$.

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

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