If y = tan^{-1} sqrt{frac{1-sin x}{1+sin x}}, | vedclass

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(A) Given $y = 	an^{-1} \sqrt{\frac{1-\sin x}{1+\sin x}}$.Using the identity $\sin x = \cos(\frac{\pi}{2}-x)$, we have:$y = 	an^{-1} \sqrt{\frac{1-\

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$-\frac{1}{2}$

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(A) Given $y = 	an^{-1} \sqrt{\frac{1-\sin x}{1+\sin x}}$.Using the identity $\sin x = \cos(\frac{\pi}{2}-x)$, we have:$y = 	an^{-1} \sqrt{\frac{1-\cos(\frac{\pi}{2}-x)}{1+\cos(\frac{\pi}{2}-x)}}$.Using half-angle formulas $1-\cos 	heta = 2\sin^2(\frac{	heta}{2})$ and $1+\cos 	heta = 2\cos^2(\frac{	heta}{2})$:$y = 	an^{-1} \sqrt{\frac{2\sin^2(\frac{\pi}{4}-\frac{x}{2})}{2\cos^2(\frac{\pi}{4}-\frac{x}{2})}} = 	an^{-1} \sqrt{	an^2(\frac{\pi}{4}-\frac{x}{2})}$.Assuming the principal value range, $y = \frac{\pi}{4} - \frac{x}{2}$.Now, differentiating with respect to $x$:$\frac{dy}{dx} = \frac{d}{dx}(\frac{\pi}{4} - \frac{x}{2}) = 0 - \frac{1}{2} = -\frac{1}{2}$.Thus, the value at $x = \frac{\pi}{6}$ is $-\frac{1}{2}$.

If $y = \tan^{-1} \sqrt{\frac{1-\sin x}{1+\sin x}}$, then the value of $\frac{dy}{dx}$ at $x = \frac{\pi}{6}$ is

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