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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

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• $1+cos\theta = 2cos^2\large\frac{\theta}{2}$
• $1-cos\theta=2sin^2\large\frac{\theta}{2}$
• $tan(A-B) = \large \frac{tanA-tanB}{1+tanAtanB}$
• $$tan\large\frac{\pi}{4}=1$$
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.