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Evaluate $\tan^{-1} \left (\large \frac {{\sqrt {1+x}}-{\sqrt {1-x}}}{{\sqrt {1+x}}+{\sqrt {1-x}}} \right )$$,\frac{-1}{\sqrt2} \leq x \leq 1$

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Toolbox:
  • $ 1+\cos\theta = 2 \cos^2\frac{\theta}{2} $
  • $ 1-\cos\theta=2 \sin^2\frac{\theta}{2}$
  • $\tan(A-B) = \large \frac{\tan A-\tan B}{1+\tan A \tan B}$
  • \(\tan\frac{\pi}{4}=1\)
Given $\tan^{-1} \left ( \large \frac {{\sqrt {1+x}}-{\sqrt {1-x}}}{{\sqrt {1+x}}+{\sqrt {1-x}}} \right ),-\frac{1}{\sqrt2} \leq x \leq 1$
Let \( x=\cos\theta\) \( \Rightarrow \theta = \cos^{-1}x\)
$\rightarrow $\(\sqrt{1+x}=\sqrt{1+\cos\theta}=\sqrt{2\cos^2\frac{\theta}{2}}=\sqrt{2}\cos\frac{\theta}{2}\)
$\rightarrow$ \(\sqrt{1-x}=\sqrt{1-\cos\theta}=\sqrt{2\sin^2\frac{\theta}{2}}=\sqrt{2}\sin\frac{\theta}{2}\)
$\rightarrow$ \( \sqrt{1+x}-\sqrt{1-x} \)\(=\sqrt{2\cos\frac{\theta}{2}}-\sqrt{2\sin\frac{\theta}{2}} =\) \( \sqrt 2\;(\cos\frac{\theta}{2}-\sin\frac{\theta}{2}) \)
$\rightarrow$ \( \sqrt{1+x}+\sqrt{1-x} \)\(=\sqrt{2\cos\frac{\theta}{2}}+\sqrt{2\sin\frac{\theta}{2}} \) \(= \sqrt 2\;(\cos\frac{\theta}{2}+\sin\frac{\theta}{2}) \)
$\Rightarrow $ \(\large \frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}= \large\frac{ \sqrt 2\;(\cos\frac{\theta}{2}-\sin\frac{\theta}{2})}{ \sqrt 2\;(\cos\frac{\theta}{2}-\sin\frac{\theta}{2})}\)
Dividing numerator and denonimator by \(\sqrt 2 \) and \(\cos\frac{\theta}{2}\), this reduces to:
$\Rightarrow $\(\;\large \frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}} \) = \(\Large \frac{1-\frac{sin\frac{\theta}{2}}{cos\frac{\theta}{2}}}{1+\frac{sin\frac{\theta}{2}}{cos\frac{\theta}{2}}}\) = \(\large \frac{1-\tan\frac{\theta}{2}}{1+\tan\frac{\theta}{2}}\)
Since \(\tan\frac{\pi}{4} = 1\) we can rewrite this as \( \large \frac{\tan\frac{\pi}{4}-\tan\frac{\theta}{2}}{1+\tan\frac{\pi}{4}.\tan\frac{\theta}{4}}\)
We know that $\tan(A-B) = \large \frac{\tan A-\tan B}{1+\tan A \tan B}$
By substituting for $A =$\(\large \frac{\pi}{4}\), \(\:B=\large\frac{\theta}{2}\), we get \(\large \frac{\tan\frac{\pi}{4}-\tan\frac{\theta}{2}}{1+\tan\frac{\pi}{4}.\tan\frac{\theta}{4}}\) \(=\tan(\frac{\pi}{4}-\frac{\theta}{2})\)
\(\Rightarrow\: \tan^{-1} \tan \big( \large \frac{\pi}{4}-\frac{\theta}{2}\)\( \big) \) \(= \large \frac{\pi}{4}-\frac{\theta}{2}\)
By substituting the value of \(\theta=\cos^{-1}x\) \(\Rightarrow\:\large \frac{\pi}{4}-\frac{1}{2}\)\(\cos^{-1}x \)
Answer: $ \tan^{-1} \left ( \large \frac {{\sqrt {1+x}}-{\sqrt {1-x}}}{{\sqrt {1+x}}+{\sqrt {1-x}}} \right ),-\frac{1}{\sqrt2} \leq x \leq 1 = $ \(\large \frac{\pi}{4}-\frac{1}{2}\)\(\cos^{-1}x \)
answered Mar 14, 2013 by balaji.thirumalai
edited Dec 19, 2013 by balaji.thirumalai
 

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