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Evaluate $\cot^{-1} \left ( \large \frac {{\sqrt {1+\sin x}}+{\sqrt {1-\sin x}}}{{\sqrt {1+\sin x}}-{\sqrt {1-\sin x}}} \right ) = \large \frac {x}{2}$$ , x \in \left ( 0, \frac{\pi}{4} \right )$

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  • \(sin^2\frac{x}{2}+cos^2\frac{x}{2}=1\)
  • \((a+b)^2=a^2+b^2+2ab\)
  • \(sinx=2sin\frac{x}{2}.cos\frac{x}{2}\)
  • \((a-b)^2=a^2+b^2-2ab\)
  • If \( x \in \bigg( 0 \frac{\pi}{4} \bigg) \) then take \( \sqrt{1-sinx} = cos\frac{x}{2}-sin\frac{x}{2}\)
  • If \(x \in \bigg( \frac{\pi}{4} \frac{\pi}{2} \bigg)\) thentake \( \sqrt{1-sinx}=sin\frac{x}{2}-cos\frac{x}{2}\)
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, \frac{\pi}{4} \right )$
We know that \(sin^2\frac{x}{2}+cos^2\frac{x}{2} = 1 \) and \(sinx=2\sin\frac{x}{2}.cos\frac{x}{2}\)
\(\Rightarrow 1+sinx=sin^2\frac{x}{2}+cos^2\frac{x}{2}+2sin\frac{x}{2}cos\frac{x}{2} = (cos\frac{x}{2}+sin\frac{x}{2})^2\)
\( \Rightarrow 1-sinx=sin^2\frac{x}{2}+cos^2\frac{x}{2}-2sin\frac{x}{2}cos\frac{x}{2} = (cos\frac{x}{2}-sin\frac{x}{2})^2\)
\(\sqrt{1+sinx}=\sqrt{(cos\frac{x}{2}+sin\frac{x}{2})^2}=cos\frac{x}{2}+sin\frac{x}{2}\)
\(\sqrt{1-sinx}=\sqrt{(cos\frac{x}{2}-sin\frac{x}{2})^2}=cos\frac{x}{2}-sin\frac{x}{2}\)
\(\sqrt{1+sinx} + \sqrt{1-sinx}= cos\frac{x}{2}+sin\frac{x}{2}+cos\frac{x}{2}-sin\frac{x}{2} = 2 cos \frac{x}{2} \)
\(\sqrt{1+sinx} - \sqrt{1-sinx}= cos\frac{x}{2}+sin\frac{x}{2}- (cos\frac{x}{2}-sin\frac{x}{2}) = 2 sin \frac{x}{2} \)
$\Rightarrow \left ( \Large \frac {{\sqrt {1+sinx}}+{\sqrt {1-sinx}}}{{\sqrt {1+sinx}}-{\sqrt {1-sinx}}} \right ) = \large \frac{2 cos \frac{x}{2}}{2 sin \frac{x}{2}} $$= cot \frac{x}{2}$
$\Rightarrow cot^{-1} \left ( \Large \frac {{\sqrt {1+sinx}}+{\sqrt {1-sinx}}}{{\sqrt {1+sinx}}-{\sqrt {1-sinx}}} \right ) =cot^{-1} cot\frac{x}{2}=\large \frac{x}{2}$
answered Mar 14, 2013 by balaji.thirumalai
 

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