If $|x| \le 1$, then $2 \tan^{-1} x + \sin^{-1} \left( \frac{2x}{1 + x^2} \right)$ is equal to

$4 \tan^{-1} x$

The correct answer is Option (1) → $4 \tan^{-1} x$ ##

We have, $2 \tan^{-1} x + \sin^{-1} \left( \frac{2x}{1 + x^2} \right)$

Let $x = \tan \theta ⇒\theta = \tan^{-1} x$

$∴2 \tan^{-1} x + \sin^{-1} \left( \frac{2x}{1 + x^2} \right) = 2 \tan^{-1}(\tan \theta) + \sin^{-1} \left( \frac{2 \tan \theta}{1 + \tan^2 \theta} \right)$

$= 2\theta + \sin^{-1}(\sin 2\theta)$

$= 2\theta + 2\theta = 4\theta = 4 \tan^{-1} x$