Question:
Prove $\tan ^{-1} \sqrt{x}=\frac{1}{2} \cos ^{-1}\left(\frac{1-x}{1+x}\right), x \in[0,1]$
Solution:
Let $x=\tan ^{2} \theta$. Then, $\sqrt{x}=\tan \theta \Rightarrow \theta=\tan ^{-1} \sqrt{x}$.
$\therefore \frac{1-x}{1+x}=\frac{1-\tan ^{2} \theta}{1+\tan ^{2} \theta}=\cos 2 \theta$
Now, we have:
R.H.S. $=\frac{1}{2} \cos ^{-1}\left(\frac{1-x}{1+x}\right)=\frac{1}{2} \cos ^{-1}(\cos 2 \theta)=\frac{1}{2} \times 2 \theta=\theta=\tan ^{-1} \sqrt{x}=$ L.H.S.
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