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Prove the following : $ \cot^{-1} \bigg[ \large\frac{\sqrt{1+sinx}+\sqrt{1-sinx}}{\sqrt{1+sinx}-\sqrt{1-sinx}} \bigg] = \frac{x}{2}, x \in \bigg( 0, \frac{\pi}{4} \bigg). $

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1 Answer

Toolbox:
  • Write \( sinx=2sin\large\frac{x}{2}\: cos\large\frac{x}{2}\)
  • \( 1=sin^2\large\frac{x}{2}+cos2\large\frac{x}{2}\)
  • Write \( \sqrt{(cos\large\frac{x}{2}-sin\large\frac{x}{2})^2}=cos\large\frac{x}{2}-sin\large\frac{x}{2}\: if\: x \bigg(0 \large\frac{\pi}{4}\bigg)\)
L.H.S \( cot^{-1} \bigg[ \large\frac{\sqrt{(cos\large\frac{x}{2}+sin\large\frac{x}{2})^2}+\sqrt{cos\large\frac{x}{2}-sin\large\frac{x}{2})^2}}{\sqrt{cos\large\frac{x}{2}+sin\large\frac{x}{2})^2}-\sqrt{cos\large\frac{x}{2}-sin\large\frac{x}{2})^2}} \bigg]\)
\( = cot^{-1} \bigg[ \frac{cos\large\frac{x}{2}}{sin\large\frac{x}{2}} \bigg]\)
\( = cot^{-1}cot\large\frac{x}{2}=\large\frac{x}{2}=R.H.S\)

 

answered Feb 28, 2013 by thanvigandhi_1
edited Mar 19, 2013 by thanvigandhi_1
 

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