For |x| lt 1 , sin(tan^-1x) is equal to

The correct answer is Option (4) → \(\displaystyle\frac{x}{\sqrt{1+x^2}}\)

Let \(\theta = \tan^{-1}x\). Then \(\tan\theta = x\) and \(-\frac{\pi}{2} \lt \theta \lt \frac{\pi}{2}\), so \(\cos\theta \gt 0\) (the principal value branch).

    \(\sec^2\theta = 1 + \tan^2\theta = 1 + x^2\)

    \(\displaystyle \cos\theta = \frac{1}{\sqrt{1 + x^2}}\)

    \(\displaystyle \sin\theta = \tan\theta\cos\theta = \frac{x}{\sqrt{1 + x^2}}\)

NCERT (2026-27 edition):
Inverse Trigonometric Functions