Single Variable Calculus, Chapter 7, 7.6, Section 7.6, Problem 12
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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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