The value of cot^{-1} left[ frac{sqrt{1 - sin | vedclass

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(D) Given expression is $y = \cot^{-1} \left[ \frac{\sqrt{1 - \sin x} + \sqrt{1 + \sin x}}{\sqrt{1 - \sin x} - \sqrt{1 + \sin x}} \right]$.Since $x \i

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$\pi - \frac{x}{2}$

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(D) Given expression is $y = \cot^{-1} \left[ \frac{\sqrt{1 - \sin x} + \sqrt{1 + \sin x}}{\sqrt{1 - \sin x} - \sqrt{1 + \sin x}} \right]$.Since $x \in (0, \frac{\pi}{2})$, we have $\frac{x}{2} \in (0, \frac{\pi}{4})$.Using $1 \pm \sin x = (\cos \frac{x}{2} \pm \sin \frac{x}{2})^2$, we get $\sqrt{1 \pm \sin x} = |\cos \frac{x}{2} \pm \sin \frac{x}{2}|$.Since $0  \sin \frac{x}{2} > 0$.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, the expression inside $\cot^{-1}$ becomes $\frac{(\cos \frac{x}{2} - \sin \frac{x}{2}) + (\cos \frac{x}{2} + \sin \frac{x}{2})}{(\cos \frac{x}{2} - \sin \frac{x}{2}) - (\cos \frac{x}{2} + \sin \frac{x}{2})} = \frac{2 \cos \frac{x}{2}}{-2 \sin \frac{x}{2}} = -\cot \frac{x}{2}$.So, $y = \cot^{-1}(-\cot \frac{x}{2}) = \pi - \cot^{-1}(\cot \frac{x}{2}) = \pi - \frac{x}{2}$.

The value of $\cot^{-1} \left[ \frac{\sqrt{1 - \sin x} + \sqrt{1 + \sin x}}{\sqrt{1 - \sin x} - \sqrt{1 + \sin x}} \right]$, where $x \in (0, \frac{\pi}{2})$ is...

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