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prove that \[tan^{-1} \left ( \frac {{\sqrt {1+x}}-{\sqrt {1-x}}}{{\sqrt {1+x}}+{\sqrt {1-x}}} \right ) = \frac {\pi}{4} -\frac{1}{2} cos^{-1}x,-\frac{1}{\sqrt2} \leq x \leq 1 \]

This question has appeared in model paper 2012

Can you answer this question?
 
 

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Toolbox:
  • $ 1+cos\theta = 2cos^2\large\frac{\theta}{2} $
  • $ 1-cos\theta=2sin^2\large\frac{\theta}{2}$
  • $tan(A-B) = \large \frac{tanA-tanB}{1+tanAtanB}$
  • \(tan\large\frac{\pi}{4}=1\)
Given $tan^{-1} \left ( \large\frac {{\sqrt {1+x}}-{\sqrt {1-x}}}{{\sqrt {1+x}}+{\sqrt {1-x}}} \right ) =\large \frac {\pi}{4} -\large\frac{1}{2} cos^{-1}x,-\large\frac{1}{\sqrt2} \leq x \leq 1$
 
Let \( x=cos\theta\) \( \Rightarrow \theta = cos^{-1}x\)
\(\sqrt{1+x}=\sqrt{1+cos\theta}=\sqrt{2cos^2\large\frac{\theta}{2}}=\sqrt{2}cos\large\frac{\theta}{2}\)
\(\sqrt{1-x}=\sqrt{1-cos\theta}=\sqrt{2sin^2\large\frac{\theta}{2}}=\sqrt{2}sin\large\frac{\theta}{2}\)
\( \sqrt{1+x}-\sqrt{1-x} \)\(=\sqrt{2}cos\large\frac{\theta}{2}-\sqrt{2}sin\large\frac{\theta}{2} =\) \( \sqrt 2\;(cos\large\frac{\theta}{2}-sin\large\frac{\theta}{2}) \)
\( \sqrt{1+x}+\sqrt{1-x} \)\(=\sqrt{2}cos\large\frac{\theta}{2}+\sqrt{2}sin\large\frac{\theta}{2} \) \(= \sqrt 2\;(cos\large\frac{\theta}{2}+sin\large\frac{\theta}{2}) \)
\(\Large \frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}= \large\frac{ \sqrt 2\;(cos\frac{\theta}{2}-sin\large\frac{\theta}{2})}{ \sqrt 2\;(cos\large\frac{\theta}{2}+sin\large\frac{\theta}{2})}\)
\(\frac{cos\frac{\theta}{2}-sin\frac{\theta}{2}}{cos\frac{\theta}{2}+sin\frac{\theta}{2}}\)
 
Dividing numerator and denonimator by  \(cos\large\frac{\theta}{2}\),
 
\( \Rightarrow \)This reduces to \(\;\Large \frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}} \) = \(\Large \frac{1-\large\frac{sin\large\frac{\theta}{2}}{cos\large\frac{\theta}{2}}}{1+\large\frac{sin\large\frac{\theta}{2}}{cos\large\frac{\theta}{2}}}\)=\(\Large \frac{1-tan\large\frac{\theta}{2}}{1+tan\large\frac{\theta}{2}}\)
 
Since \(1=tan\large\frac{\pi}{4}\) we can rewrite this as \( \large \frac{tan\large\frac{\pi}{4}-tan\large\frac{\theta}{2}}{1+tan\large\frac{\pi}{4}.tan\large\frac{\theta}{4}}\)
We know that $tan(A-B) = \large \frac{tanA-tanB}{1+tanAtanB}$
 
By substituting for $A =$\(\large\frac{\pi}{4}, \:B=\large\frac{\theta}{2}\), we get \(\large \frac{tan\frac{\pi}{4}-tan\large\frac{\theta}{2}}{1+tan\frac{\pi}{4}.tan\large\frac{\theta}{4}}\)\(=tan(\large\frac{\pi}{4}-\large\frac{\theta}{2})\)
\(\Rightarrow\: tan^{-1}tan \bigg( \large\frac{\pi}{4}-\large\frac{\theta}{2} \bigg) = \frac{\pi}{4}-\large\frac{\theta}{2}\)
 
By substituting the value of \(\theta=cos^{-1}x\) \(\Rightarrow\:\large\frac{\pi}{4}-\large\frac{1}{2} cos^{-1}x \) = R.H.S.

 

answered Feb 23, 2013 by thanvigandhi_1
edited Mar 16, 2013 by rvidyagovindarajan_1
 

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