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प्रश 9. $$\cot ^{-1}\left[\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right]=\frac{x}{2}$$,
$$
\begin{aligned}
& x \in\left(0, \frac{\pi}{4}\right) \\
& \text { हल : L.H.S. }=\cot ^{-1}\left[\frac{(\sqrt{1+\sin x}+\sqrt{1-\sin x})}{(\sqrt{1+\sin x}-\sqrt{1-\sin x})}\right] \\
& \times \frac{(\sqrt{1+\sin x}+\sqrt{1-\sin x})}{(\sqrt{1+\sin x}+\sqrt{1-\sin x})}
\end{aligned}
$$
$$
\begin{aligned}
& =\cot ^{-1}\left[\frac{2+2 \sqrt{1-\sin ^2 x}}{2 \sin x}\right] \\
& =\cot ^{-1}\left[\frac{2+2 \cos x}{2 \sin x}\right] \\
& =\cot ^{-1}\left[\frac{1+\cos x}{\sin x}\right] \\
& =\cot ^{-1}\left[\frac{1+2 \cos ^2 \frac{x}{2}-1}{2 \sin \frac{x}{2} \cos \frac{x}{2}}\right] \text {. } \\
& =\cot ^{-1}\left[\frac{2 \cos ^2 \frac{x}{2}}{2 \sin \frac{x}{2} \cos \frac{x}{2}}\right] \\
& =\cot ^{-1}\left[\frac{\cos \frac{x}{2}}{\sin \frac{x}{2}}\right]=\cot ^{-1}\left[\cot \frac{x}{2}\right] \\
& =\frac{x}{2}=\text { R.H.S. }
\end{aligned}
$$
इति सिद्धम्।
अत: L.H.S. = R.H.S.
