for $|x| < 1$, $sin (tan^{-1}x)$ equal to

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

By definition, $\tan \theta = x$

This can be written as:

$\tan \theta = \frac{x}{1}$

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Construct a right triangle where:

Opposite side $= x$

Adjacent side $= 1$

This gives $\tan \theta = \frac{\text{opposite}}{\text{adjacent}} = \frac{x}{1} = x$

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Using the Pythagorean theorem:

$\text{Hypotenuse} = \sqrt{x^2 + 1^2}$

$\text{Hypotenuse} = \sqrt{x^2 + 1}$

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For the sine ratio:

$\sin \theta = \frac{\text{opposite}}{\text{hypotenuse}}$

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

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Since $\theta = \tan^{-1}x$:

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