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

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

Let $\theta=\tan^{-1}(x)$.

$\tan\theta=\frac{x}{1}$ gives a right triangle with:

Opposite $=x$, Adjacent $=1$, Hypotenuse $=\sqrt{1+x^{2}}$.

Hence,

$\sin\theta=\frac{x}{\sqrt{1+x^{2}}}$

$\sin(\tan^{-1}x)=\frac{x}{\sqrt{1+x^{2}}}$