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Prove that $\tan ^{-1} \sqrt{x}=\frac{1}{2} \cos ^{-1}\left(\frac{1-x}{1+x}\right)$, $x \in[0,1]$
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Put $x=\tan ^2 \theta, \quad \theta=\tan ^{-1} \sqrt{x}$
R.H.S. $=\frac{1}{2} \cos ^{-1}\left(\frac{1-x}{1+x}\right)$
$=\frac{1}{2} \cos ^{-1}\left(\frac{1-\tan ^2 \theta}{1+\tan ^2 \theta}\right)=\frac{1}{2} \cos ^{-1}(\cos 2 \theta)=\theta$
$=\tan ^{-1} \sqrt{x}=\frac{1}{2} \cos ^{-1}\left(\frac{1-x}{1+x}\right)$
R.H.S. $=\frac{1}{2} \cos ^{-1}\left(\frac{1-x}{1+x}\right)$
$=\frac{1}{2} \cos ^{-1}\left(\frac{1-\tan ^2 \theta}{1+\tan ^2 \theta}\right)=\frac{1}{2} \cos ^{-1}(\cos 2 \theta)=\theta$
$=\tan ^{-1} \sqrt{x}=\frac{1}{2} \cos ^{-1}\left(\frac{1-x}{1+x}\right)$
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