Simplify the expression $\tan \left( \sin^{-1} (x) \right)$

If we let the values of right triangle be...

We know that $\displaystyle \sin \theta = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{x}{1} = x$

$\theta = \sin^{-1}(x)$

And,

$\displaystyle \tan \theta = \frac{\text{opposite}}{\text{adjacent}} = \frac{x}{\sqrt{1-x^2}}$

Thus, $\displaystyle \tan \left( \sin^{-1}(x) \right) = \tan (\theta) = \frac{x}{\sqrt{1-x^2}}$

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