Answer
$cos~(2~tan^{-1}~\frac{3}{u}) = \frac{u^2-9}{u^2+9}$
Work Step by Step
Let $~~\theta = tan^{-1}~\frac{3}{u}$
Then $~~tan~\theta = \frac{3}{u}$
We can find an expression for $sin~\theta$:
$sin~\theta = \frac{3}{\sqrt{u^2+3^2}}$
$sin~\theta = \frac{3}{\sqrt{u^2+9}}$
We can find $cos~(2\theta)$:
$cos~(2\theta) = 1-2~sin^2~\theta$
$cos~(2\theta) = 1-(2)(\frac{3}{\sqrt{u^2+9}})^2$
$cos~(2\theta) = 1-\frac{18}{u^2+9}$
$cos~(2\theta) = \frac{u^2-9}{u^2+9}$
Therefore, $~~cos~(2~tan^{-1}~\frac{3}{u}) = \frac{u^2-9}{u^2+9}$
