Answer:
LHS \[={{\cot }^{-\,1}}\left[ \frac{\sqrt{1+\sin \,x}+\sqrt{1-\sin x}}{\sqrt{1+\sin \,x}-\sqrt{1-\sin x}} \right]\] \[={{\cot }^{-\,1}}\left[ \frac{\begin{align} & \sqrt{{{\sin }^{2}}\frac{x}{2}+{{\cos }^{2}}\frac{x}{2}+2\sin \frac{x}{2}\cos \frac{x}{2}} \\ & +\,\sqrt{{{\sin }^{2}}\frac{x}{2}+{{\cos }^{2}}\frac{x}{2}-2\sin \frac{x}{2}\cos \frac{x}{2}} \\ \end{align}}{\begin{align} & \sqrt{{{\sin }^{2}}\frac{x}{2}+{{\cos }^{2}}\frac{x}{2}+2\sin \frac{x}{2}\cos \frac{x}{2}} \\ & -\,\sqrt{{{\sin }^{2}}\frac{x}{2}+{{\cos }^{2}}\frac{x}{2}-2\sin \frac{x}{2}\cos \frac{x}{2}} \\ \end{align}} \right]\] \[\left[ \because \,\,1={{\sin }^{2}}\frac{x}{2}+{{\cos }^{2}}\frac{x}{2}\,\,and\,\,\sin x=2\sin \frac{x}{2}\cos \frac{x}{2} \right]\] \[={{\cot }^{-\,1}}\left[ \frac{\sqrt{{{\left( \cos \frac{x}{2}+sin\frac{x}{2} \right)}^{2}}}+\sqrt{{{\left( \cos \frac{x}{2}-sin\frac{x}{2} \right)}^{2}}}}{\sqrt{{{\left( \cos \frac{x}{2}+sin\frac{x}{2} \right)}^{2}}}-\sqrt{{{\left( \cos \frac{x}{2}-sin\frac{x}{2} \right)}^{2}}}} \right]\] \[[\because \,\,\,{{a}^{2}}+{{b}^{2}}\pm 2ab={{(a\pm b)}^{2}}]\] \[={{\cot }^{-\,1}}\left[ \frac{\cos \frac{x}{2}+\sin \frac{x}{2}+\cos \frac{x}{2}-\sin \frac{x}{2}}{\cos \frac{x}{2}+\sin \frac{x}{2}-\cos \frac{x}{2}+\sin \frac{x}{2}} \right]\] \[={{\cot }^{-\,1}}\left[ \frac{2\cos \frac{x}{2}}{2\sin \frac{x}{2}} \right]\] \[={{\cot }^{-\,1}}\left( \cot \frac{x}{2} \right)=\frac{x}{2}=RHS\] \[\left[ \because \,\,\,0<\frac{x}{2}<\frac{\pi }{8} \right]\] Hence proved.
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