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Prove that $\cos\bigg\{2\cot^{-1}\sqrt{\large\frac{1-x}{1+x}}\bigg\}+x=0$

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Toolbox:
  • put \( x=cos2\theta \Rightarrow 2\theta=cos^{-1}x\)
  • \( 1-cos2\theta=2\: sin^2\theta\)
  • \( 1+cos2\theta = 2cos^2\theta \)
  • \(cot^{-1}\: tan\theta = \large\frac{\pi}{2}-\theta \)
  • \(cos(\pi-x)=-cosx\)
By taking \(x=cos2\theta,\:\:\sqrt{\large\frac{1-x}{1+x}}=\sqrt{\large\frac{1-cos2\theta}{1+cos2\theta}}\)
\(=\sqrt{\large\frac{2sin^2\theta}{2cos^2\theta}}=\frac{sin\theta}{cos\theta}=tan\theta\)
\(\Rightarrow\:cot^{-1}\sqrt{\large\frac{1-x}{1+x}}=cot^{-1}tan\theta\)
 
Substituting the value of x in the given eqn. we get
\( cos \{ 2 \: cot^{-1}\: tan\theta \} + cos2\theta \)
 
\(\Rightarrow\: cos \{ 2 ( \large\frac{\pi}{2} - \theta ) \} + cos2\theta \)
\(=cos(\pi-2\theta)+cos2\theta\)
\( = -cos2\theta+cos2\theta=0\)

 

answered Feb 20, 2013 by thanvigandhi_1
edited Mar 19, 2013 by thanvigandhi_1
 

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