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