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

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Question

(C) Let $y = \cot ^{-1}\left[\frac{\sqrt{1-\sin x}+\sqrt{1+\sin x}}{\sqrt{1-\sin x}-\sqrt{1+\sin x}}\right]$.Rationalizing the expression inside the b

Vedclass · Accepted Answer

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

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(C) Let $y = \cot ^{-1}\left[\frac{\sqrt{1-\sin x}+\sqrt{1+\sin x}}{\sqrt{1-\sin x}-\sqrt{1+\sin x}}\right]$.Rationalizing the expression inside the bracket:$\frac{(\sqrt{1-\sin x}+\sqrt{1+\sin x})^2}{(1-\sin x)-(1+\sin x)} = \frac{1-\sin x + 1+\sin x + 2\sqrt{(1-\sin x)(1+\sin x)}}{-2\sin x} = \frac{2 + 2\sqrt{1-\sin^2 x}}{-2\sin x} = \frac{2 + 2\cos x}{-2\sin x} = -\frac{1+\cos x}{\sin x}$.Using half-angle identities $1+\cos x = 2\cos^2\frac{x}{2}$ and $\sin x = 2\sin\frac{x}{2}\cos\frac{x}{2}$:$-\frac{2\cos^2\frac{x}{2}}{2\sin\frac{x}{2}\cos\frac{x}{2}} = -\cot\frac{x}{2}$.Thus,$y = \cot^{-1}(-\cot\frac{x}{2})$.Since $\cot^{-1}(-z) = \pi - \cot^{-1}(z)$,we have $y = \pi - \cot^{-1}(\cot\frac{x}{2})$.Given $x \in (0, \frac{\pi}{4})$,then $\frac{x}{2} \in (0, \frac{\pi}{8})$,so $y = \pi - \frac{x}{2}$.

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 $x \in\left(0, \frac{\pi}{4}\right)$ is

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