Prove the following $\cot ^{-1}\left[\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right]=\frac{x}{2}, x \in\left(0, \frac{\pi}{4}\right)$
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Prove the following $\cot ^{-1}\left[\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right]=\frac{x}{2}, x \in\left(0, \frac{\pi}{4}\right)$

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SOLUTION :  L.H.S. $=\cot ^{-1}\left[\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right]$

$=\cot ^{-1}\left(\frac{\sqrt{\sin ^2 \frac{x}{2}+\cos ^2 \frac{x}{2}+2 \sin \frac{x}{2} \cos \frac{x}{2}}+\sqrt{\cos ^2 \frac{x}{2}+\sin ^2 \frac{x}{2}-2 \sin \frac{x}{2} \cos \frac{x}{2}}}{\sqrt{\sin ^2 \frac{x}{2}+\cos ^2 \frac{x}{2}+2 \sin \frac{x}{2} \cos \frac{x}{2}-\sqrt{\sin ^2 \frac{x}{2}+\cos ^2 \frac{x}{2}-2 \sin \frac{x}{2} \cos \frac{x}{2}}}}\right) \\$

$\because\left[\sin ^2 \frac{x}{2}+\cos \frac{x}{2}\right]=1 \text { and }\left[\sin x=2 \sin \frac{x}{2} \cdot \cos ^2 \frac{x}{2}\right]=\cot ^{-1}\left[\frac{\sqrt{\left(\sin \frac{x}{2}+\cos \frac{x}{2}\right)^2+\sqrt{\left(\cos \frac{x}{2}-\sin \frac{x}{2}\right)^2}}}{\sqrt{\left(\sin \frac{x}{2}+\cos \frac{x}{2}\right)^2-\sqrt{\left(\cos \frac{x}{2}-\sin \frac{x}{2}\right)^2}}}\right. \\$

$=\cot ^{-1}\left[\frac{\sin \frac{x}{2}+\cos \frac{x}{2}+\cos \frac{x}{2}-\sin \frac{x}{2}}{\sin \frac{x}{2}+\cos \frac{x}{2}-\cos \frac{x}{2}+\sin \frac{x}{2}}\right]=\cot ^{-1}\left[\frac{2 \cos \frac{x}{2}}{2 \sin \frac{x}{2}}\right]=\cot ^{-1} \cdot \cot \frac{x}{2}=\frac{x}{2}=\text { R.H.S. } \\$

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