## Trigonometry (11th Edition) Clone

$tan~(arccos~u) = \frac{u~\sqrt{1^2-u^2}}{1^2-u^2}$
$\theta = arcsin(u)$ $sin~\theta = u = \frac{u}{1} = \frac{opposite}{hypotenuse}$ 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, $tan~(arccos~u) = \frac{u~\sqrt{1^2-u^2}}{1^2-u^2}$