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