Assertion (A): For $-1 < x < 1$, $\frac{d}{dx}\tan^{-1}\left(\frac{x}{\sqrt{1 - x^{2}}}\right) = \frac{1}{\sqrt{1 - x^{2}}}$.

Reason (R): For $-1 < x < 1$, $\tan^{-1}\left(\frac{x}{\sqrt{1 - x^{2}}}\right) = \sin^{-1} x$.

Explanation

1. 01

   R: put $x = \sin\theta$ with $-\frac{\pi}{2} < \theta < \frac{\pi}{2}$; then $\sqrt{1 - x^{2}} = \cos\theta > 0$, the fraction is $\tan\theta$, and the function is $\theta = \sin^{-1} x$. True.

2. 02

   A follows at once: the derivative of $\sin^{-1} x$ is $\frac{1}{\sqrt{1 - x^{2}}}$. Both are true, and R explains A.