Differentiate the function with respect to x: | vedclass

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(C) Let $y = \cot ^{-1}\left[\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right]$.We know that $1+\sin x = \cos^2 \frac{x}{

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$\frac{1}{2}$

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(C) Let $y = \cot ^{-1}\left[\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right]$.We know that $1+\sin x = \cos^2 \frac{x}{2} + \sin^2 \frac{x}{2} + 2\sin \frac{x}{2} \cos \frac{x}{2} = (\cos \frac{x}{2} + \sin \frac{x}{2})^2$.Similarly,$1-\sin x = (\cos \frac{x}{2} - \sin \frac{x}{2})^2$.Since $0  \sin \frac{x}{2}$.Thus,$\sqrt{1+\sin x} = \cos \frac{x}{2} + \sin \frac{x}{2}$ and $\sqrt{1-\sin x} = \cos \frac{x}{2} - \sin \frac{x}{2}$.Substituting these into the expression:$y = \cot ^{-1}\left[\frac{(\cos \frac{x}{2} + \sin \frac{x}{2}) + (\cos \frac{x}{2} - \sin \frac{x}{2})}{(\cos \frac{x}{2} + \sin \frac{x}{2}) - (\cos \frac{x}{2} - \sin \frac{x}{2})}\right]$$y = \cot ^{-1}\left[\frac{2\cos \frac{x}{2}}{2\sin \frac{x}{2}}\right]$$y = \cot ^{-1}(\cot \frac{x}{2}) = \frac{x}{2}$.Therefore,$\frac{dy}{dx} = \frac{d}{dx}(\frac{x}{2}) = \frac{1}{2}$.

Differentiate the function with respect to $x$: $\cot ^{-1}\left[\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right]$,where $0 < x < \frac{\pi}{2}$.

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