Answer
$tan~(arccos~u) = \frac{u~\sqrt{1^2-u^2}}{1^2-u^2}$
Work Step by Step
$\theta = arcsin(u)$
$sin~\theta = u = \frac{u}{1} = \frac{opposite}{hypotenuse}$
Note that $\theta$ is in quadrant I since $u \gt 0$. We can find the value of the adjacent side:
$adjacent = \sqrt{1^2-u^2}$
We can find the value of $tan~\theta$:
$tan~\theta = \frac{opposite}{adjacent}$
$tan~\theta = \frac{u}{\sqrt{1^2-u^2}}$
$tan~\theta = \frac{u}{\sqrt{1^2-u^2}}~\frac{\sqrt{1^2-u^2}}{\sqrt{1^2-u^2}}$
$tan~\theta = \frac{u~\sqrt{1^2-u^2}}{1^2-u^2}$
Therefore, $tan~(arccos~u) = \frac{u~\sqrt{1^2-u^2}}{1^2-u^2}$
