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

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

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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 $1+\sin x = (\cos \frac{x}{2} + \sin \frac{x}{2})^2$ and $1-\sin x = (\cos \frac{x}{2} - \sin \frac{x}{2})^2$,we get:$y = 	an^{-1} \left( \sqrt{\frac{(\cos \frac{x}{2} + \sin \frac{x}{2})^2}{(\cos \frac{x}{2} - \sin \frac{x}{2})^2}} ight) = 	an^{-1} \left( \frac{\cos \frac{x}{2} + \sin \frac{x}{2}}{\cos \frac{x}{2} - \sin \frac{x}{2}} ight)$.Dividing numerator and denominator by $\cos \frac{x}{2}$,we get:$y = 	an^{-1} \left( \frac{1 + 	an \frac{x}{2}}{1 - 	an \frac{x}{2}} ight) = 	an^{-1} \left( 	an(\frac{\pi}{4} + \frac{x}{2}) ight)$.Since $0 \leq x 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,at $x = \frac{\pi}{6}$,$\frac{dy}{dx} = \frac{1}{2}$.

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

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