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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 by thanvigandhi_1
edited Mar 15, 2013 by thanvigandhi_1
 

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