Answer
$cos~(tan^{-1}~(\frac{4}{5})) = \frac{5\sqrt{41}}{41}$
Work Step by Step
$\theta = tan^{-1}(\frac{4}{5})$
$tan~\theta = \frac{4}{5} = \frac{opposite}{adjacent}$
Note that $\theta$ is in quadrant I. We can find the value of the hypotenuse:
$hypotenuse = \sqrt{4^2+5^2} = \sqrt{41}$
We can find the value of $cos~\theta$:
$cos~\theta = \frac{adjacent}{hypotenuse}$
$cos~\theta = \frac{5}{\sqrt{41}}$
$cos~\theta = \frac{5}{\sqrt{41}}~\frac{\sqrt{41}}{\sqrt{41}}$
$cos~\theta = \frac{5\sqrt{41}}{41}$
Therefore, $cos~(tan^{-1}~(\frac{4}{5})) = \frac{5\sqrt{41}}{41}$
