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# Prove that $tan^{-1} \sqrt{x} = \frac{1}{2} cos^{-1} \;\; \bigg(\frac{1-x}{1+x}\bigg), x\in [0,1]$

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• $\large\frac{1-tan^2\theta}{1+tan^2\theta}$$=cos2\theta$
Given $tan^{-1} \sqrt{x} =\large \frac{1}{2} cos^{-1} \;\; \bigg(\large\frac{1-x}{1+x}\bigg), x\in [0,1]$

We know that $\large\frac{1-tan^2\theta}{1+tan^2\theta}$$=cos2\theta$

Let $tan \theta = \sqrt x \Rightarrow \theta = tan^{-1}\sqrt x\;$ and $\;x = tan^2 \theta$

Given, $\;x = tan^2 \theta \rightarrow 1 - x = 1 - tan^\theta\;$ and $1 + x = 1 + tan^2\theta$
$\large \frac{1-x}{1+x}$$=\large \frac{1-tan^2\theta}{1+tan^2\theta}$ which as we know is $=cos2\theta$

$\Rightarrow \large\frac{1}{2} cos^{-1}\big(\large\frac{1-x}{1+x}\big)=\large\frac{1}{2}cos^{-1} \bigg( \large\frac{1-tan^2\theta}{1+tan^2\theta} \bigg)$ $= \large\frac{1}{2}cos^{-1}\: cos2\theta =\large\frac{1}{2}.2\theta= \theta$

Since $\theta = tan^{-1}\sqrt x$, L.H.S. = R.H.S.

answered Feb 23, 2013
edited Mar 15, 2013