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If $y=\tan ^{-1}\left(\sqrt{\frac{1+\sin x}{1-\sin x}}\right), 0 \leq x < \frac{\pi}{2}$, then $\frac{d y}{d x}$ at $x=\frac{\pi}{6}$ is
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$\frac{1}{2}$
$\begin{aligned} & y=\tan ^{-1}\left(\sqrt{\frac{1+\sin x}{1-\sin x}}\right) \\ & =\tan ^{-1}\left[\sqrt{\frac{\left(\cos \frac{x}{2}+\sin \frac{x}{2}\right)^2}{\left(\cos \frac{x}{2}-\sin \frac{x}{2}\right)^2}}\right]=\tan ^{-1}\left[\frac{\left(\cos \frac{x}{2}+\sin \frac{x}{2}\right)^2}{\left(\cos \frac{x}{2}-\sin \frac{x}{2}\right)^2}\right] \\ & =\tan ^{-1}\left(\frac{1+\tan \frac{x}{2}}{1-\tan \frac{x}{2}}\right) \\ & =\tan ^{-1}\left(\frac{\tan \frac{\pi}{4}+\tan \frac{x}{2}}{1-\tan \frac{\pi}{4} \cdot \tan \frac{x}{2}}\right) \\ & =\tan ^{-1}\left[\tan \left(\frac{\pi}{4}+\frac{x}{2}\right)\right]=\frac{\pi}{4}+\frac{x}{2} \\ & \therefore \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1}{2} \\ & \end{aligned}$
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