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

This question is Q.No.9 of Misc. chapter 2

Toolbox:
• $$\large\frac{1-tan^2\theta}{1+tan^2\theta}$$$$=cos2\theta$$
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 18, 2013