question_answer

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

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.