$\mathrm{for}\left|x\right|<1,\sin \left({\tan }^{-1}x\right)\mathrm{is}\mathrm{equal}\mathrm{to}$

The correct option is:
$\frac{x}{\sqrt{1+x^2}}$

Step-by-Step Derivation:

To evaluate the trigonometric expression, we can use a substitution method. Let us define a new variable:
$\theta ={\tan }^{-1}\left(x\right)$

By the definition of the inverse tangent function, this is equivalent to:
$\tan \theta =x$
where
$-\frac{\pi }{2}<\theta <\frac{\pi }{2}$

Our goal is to find the value of:
$\sin \left({\tan }^{-1}x\right)=\sin \theta$

We can relate $\sin \theta$ to $\tan \theta$ using the fundamental trigonometric identity:
$\sin \theta =\frac{\tan \theta }{\sec \theta }$

Next, we express $\sec \theta$ in terms of $\tan \theta$ using the Pythagorean identity:
${\sec }^2\theta =1+{\tan }^2\theta$

Taking the square root of both sides, and noting that $\sec \theta$ is positive in the interval $(-\frac{\pi }{2},\frac{\pi }{2})$, we have:
$\sec \theta =\sqrt{1+{\tan }^2\theta }$

Substituting $\tan \theta =x$ into the equations:
$\sec \theta =\sqrt{1+x^2}$

Now, substitute these expressions back into the relation for $\sin \theta$:
$\sin \theta =\frac{x}{\sqrt{1+x^2}}$

Replacing $\theta$ with ${\tan }^{-1}x$ gives the final simplified result:
$\sin \left({\tan }^{-1}x\right)=\frac{x}{\sqrt{1+x^2}}$