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

Let $x = \tan^2 \theta$.
Then $\sqrt{x} = \tan \theta$,which implies $\theta = \tan^{-1} \sqrt{x}$.
Consider the expression $\frac{1-x}{1+x}$. Substituting $x = \tan^2 \theta$,we get:
$\frac{1 - \tan^2 \theta}{1 + \tan^2 \theta} = \cos(2\theta)$.
Now,consider the Right Hand Side $(RHS)$:
$RHS = \frac{1}{2} \cos^{-1} \left( \frac{1-x}{1+x} \right)$
$= \frac{1}{2} \cos^{-1} (\cos 2\theta)$
$= \frac{1}{2} \times 2\theta = \theta$.
Since $\theta = \tan^{-1} \sqrt{x}$,we have $RHS = \tan^{-1} \sqrt{x} = LHS$.
Hence,the identity is proved.