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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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The correct answer is:
$\pi-\frac{x}{2}$
Here,
$\cot ^{-1}\left[\frac{\sqrt{1-\sin x}+\sqrt{1+\sin x}}{\sqrt{1-\sin x}-\sqrt{1-\sin x}}\right]$
$\begin{aligned} & =\cot ^{-1}\left[\frac{(\sqrt{1-\sin x}+\sqrt{1+\sin x})^2}{(1-\sin x)-(1+\sin x)}\right] \\ & =\cot ^{-1}\left(\frac{1-\sin x+1+\sin x+2 \sqrt{1-\sin ^2 x}}{1-\sin x-1-\sin x}\right) \\ & \Rightarrow \cot ^{-1}\left(\frac{2+2 \sqrt{\cos ^2 x}}{-2 \sin x}\right) \\ & \Rightarrow \cot ^1\left(\frac{1+\cos x}{-\sin x}\right)\end{aligned}$
$\begin{gathered}\Rightarrow \cot ^{-1}\left(\frac{1+2 \cos ^2 \frac{x}{2}-1}{-2 \sin \frac{x}{2} \cos \frac{x}{2}}\right) \\ \Rightarrow \cot ^1\left(\frac{-\cos \frac{x}{2}}{\sin \frac{x}{2}}\right) \\ \Rightarrow \cot ^{-1}\left(-\cot \frac{x}{2}\right)=\cot ^{-1}\left(\cot \left(\pi-\frac{x}{2}\right)\right) \\ =\pi-\frac{x}{2}\end{gathered}$
$\cot ^{-1}\left[\frac{\sqrt{1-\sin x}+\sqrt{1+\sin x}}{\sqrt{1-\sin x}-\sqrt{1-\sin x}}\right]$
$\begin{aligned} & =\cot ^{-1}\left[\frac{(\sqrt{1-\sin x}+\sqrt{1+\sin x})^2}{(1-\sin x)-(1+\sin x)}\right] \\ & =\cot ^{-1}\left(\frac{1-\sin x+1+\sin x+2 \sqrt{1-\sin ^2 x}}{1-\sin x-1-\sin x}\right) \\ & \Rightarrow \cot ^{-1}\left(\frac{2+2 \sqrt{\cos ^2 x}}{-2 \sin x}\right) \\ & \Rightarrow \cot ^1\left(\frac{1+\cos x}{-\sin x}\right)\end{aligned}$
$\begin{gathered}\Rightarrow \cot ^{-1}\left(\frac{1+2 \cos ^2 \frac{x}{2}-1}{-2 \sin \frac{x}{2} \cos \frac{x}{2}}\right) \\ \Rightarrow \cot ^1\left(\frac{-\cos \frac{x}{2}}{\sin \frac{x}{2}}\right) \\ \Rightarrow \cot ^{-1}\left(-\cot \frac{x}{2}\right)=\cot ^{-1}\left(\cot \left(\pi-\frac{x}{2}\right)\right) \\ =\pi-\frac{x}{2}\end{gathered}$
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