# 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.

edited Mar 15, 2013