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# Prove that $cot^{-1} \left ( \frac {{\sqrt {1+sinx}}+{\sqrt {1-sinx}}}{{\sqrt {1+sinx}}-{\sqrt {1-sinx}}} \right ) = \frac {x}{2} , x \in \left ( 0, \frac{\pi}{4} \right )$

Toolbox:
• $sin^2\large\frac{x}{2}+cos^2\large\frac{x}{2}=1$
• $(a+b)^2=a^2+b^2+2ab$
• $sinx=2sin\large\frac{x}{2}.cos\large\frac{x}{2}$
• $(a-b)^2=a^2+b^2-2ab$
• If $x \in \bigg( 0 \large\frac{\pi}{4} \bigg)$ then take $\sqrt{1-sinx} = cos\large\frac{x}{2}-sin\large\frac{x}{2}$
• If $x \in \bigg( \large\frac{\pi}{4} \large\frac{\pi}{2} \bigg)$ then take $\sqrt{1-sinx}=sin\large\frac{x}{2}-cos\large\frac{x}{2}$
Given $cot^{-1} \left ( \Large \frac {{\sqrt {1+sinx}}+{\sqrt {1-sinx}}}{{\sqrt {1+sinx}}-{\sqrt {1-sinx}}} \right ) = \large \frac {x}{2}$$, x \in \left ( 0,\large \frac{\pi}{4} \right ) We know that $sin^2\large\frac{x}{2}+cos^2\large\frac{x}{2} = 1$ and $sinx=2\sin\large\frac{x}{2}.cos\large\frac{x}{2}$ $\Rightarrow 1+sinx=sin^2\large\frac{x}{2}+cos^2\large\frac{x}{2}+2sin\large\frac{x}{2}cos\large\frac{x}{2} = (cos\large\frac{x}{2}+sin\large\frac{x}{2})^2$ $\Rightarrow 1-sinx=sin^2\large\frac{x}{2}+cos^2\large\frac{x}{2}-2sin\large\frac{x}{2}cos\frac{x}{2} = (cos\large\frac{x}{2}-sin\large\frac{x}{2})^2$ $\sqrt{1+sinx}=\sqrt{(cos\large\frac{x}{2}+sin\large\frac{x}{2})^2}=cos\large\frac{x}{2}+sin\large\frac{x}{2}$ $\sqrt{1-sinx}=\sqrt{(cos\large\frac{x}{2}-sin\large\frac{x}{2})^2}=cos\large\frac{x}{2}-sin\large\frac{x}{2}$ $\sqrt{1+sinx} + \sqrt{1-sinx}= cos\large\frac{x}{2}+sin\large\frac{x}{2}+cos\large\frac{x}{2}-sin\large\frac{x}{2} = 2 cos\large \frac{x}{2}$ $\sqrt{1+sinx} - \sqrt{1-sinx}= cos\large\frac{x}{2}+sin\large\frac{x}{2}- (cos\large\frac{x}{2}-sin\large\frac{x}{2}) = 2 sin \large\frac{x}{2}$ \Rightarrow \left ( \Large \frac {{\sqrt {1+sinx}}+{\sqrt {1-sinx}}}{{\sqrt {1+sinx}}-{\sqrt {1-sinx}}} \right ) = \large \frac{2 cos \large\frac{x}{2}}{2 sin\large \frac{x}{2}}$$= cot \frac{x}{2}$

$\Rightarrow cot^{-1} \left ( \Large \frac {{\sqrt {1+sinx}}+{\sqrt {1-sinx}}}{{\sqrt {1+sinx}}-{\sqrt {1-sinx}}} \right ) =cot^{-1} cot\large\frac{x}{2}=\large \frac{x}{2}$

edited Mar 15, 2013