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

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  • \(sin^2\large\frac{x}{2}+cos^2\large\frac{x}{2}=1\)
  • \((a+b)^2=a^2+b^2+2ab\)
  • \(sinx=2sin\large\frac{x}{2}.cos\large\frac{x}{2}\)
  • \((a-b)^2=a^2+b^2-2ab\)
  • 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}$
 
$\Rightarrow cot^{-1} \left ( \Large \frac {{\sqrt {1+sinx}}+{\sqrt {1-sinx}}}{{\sqrt {1+sinx}}-{\sqrt {1-sinx}}} \right ) =cot^{-1} cot\large\frac{x}{2}=\large \frac{x}{2}$

 

answered Feb 23, 2013 by thanvigandhi_1
edited Mar 15, 2013 by thanvigandhi_1
 

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