Answer
$tan~(arccos~\frac{3}{4}) = \frac{\sqrt{7}}{3}$
Work Step by Step
$\theta = arccos(\frac{3}{4})$
$cos~\theta = \frac{3}{4} = \frac{adjacent}{hypotenuse}$
Note that $\theta$ is in quadrant I. We can find the value of the opposite side:
$opposite = \sqrt{4^2-3^2} = \sqrt{7}$
We can find the value of $tan~\theta$:
$tan~\theta = \frac{opposite}{adjacent}$
$tan~\theta = \frac{\sqrt{7}}{3}$
Therefore, $tan~(arccos~\frac{3}{4}) = \frac{\sqrt{7}}{3}$
