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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

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.

edited Mar 16, 2013