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

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(D) Given $y = 	an^{-1}\left(\sqrt{\frac{1+\sin x}{1-\sin x}}ight)$.Using the identities $1+\sin x = \left(\cos\frac{x}{2} + \sin\frac{x}{2}ight)

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$\frac{1}{2}$

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(D) Given $y = 	an^{-1}\left(\sqrt{\frac{1+\sin x}{1-\sin x}}ight)$.Using the identities $1+\sin x = \left(\cos\frac{x}{2} + \sin\frac{x}{2}ight)^2$ and $1-\sin x = \left(\cos\frac{x}{2} - \sin\frac{x}{2}ight)^2$,we have:$y = 	an^{-1}\left(\sqrt{\frac{(\cos(x/2) + \sin(x/2))^2}{(\cos(x/2) - \sin(x/2))^2}}ight) = 	an^{-1}\left(\frac{\cos(x/2) + \sin(x/2)}{\cos(x/2) - \sin(x/2)}ight)$.Dividing numerator and denominator by $\cos(x/2)$,we get:$y = 	an^{-1}\left(\frac{1 + 	an(x/2)}{1 - 	an(x/2)}ight) = 	an^{-1}\left(	an\left(\frac{\pi}{4} + \frac{x}{2}ight)ight)$.Since $0 \leqslant x Differentiating with respect to $x$,we get $y' = \frac{1}{2}$.Thus,$y'\left(\frac{\pi}{6}ight) = \frac{1}{2}$.

If $y = \tan^{-1}\left(\sqrt{\frac{1+\sin x}{1-\sin x}}\right)$,where $0 \leqslant x < \frac{\pi}{2}$,then find the value of $y'\left(\frac{\pi}{6}\right)$.

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