Answer
$sec~(arccot~\frac{\sqrt{4-u^2}}{u}) = \frac{2~\sqrt{4-u^2}}{4-u^2}$
Work Step by Step
Let $~~\theta = arccot~\frac{\sqrt{4-u^2}}{u}$
Then $~~cot~\theta = \frac{\sqrt{4-u^2}}{u}$
We can find an expression for $sec~\theta$:
$sec~\theta = \frac{\sqrt{(\sqrt{4-u^2})^2+u^2}}{\sqrt{4-u^2}}$
$sec~\theta = \frac{\sqrt{4-u^2+u^2}}{\sqrt{4-u^2}}$
$sec~\theta = \frac{\sqrt{4}}{\sqrt{4-u^2}}$
$sec~\theta = \frac{2}{\sqrt{4-u^2}}$
$sec~\theta = (\frac{2}{\sqrt{4-u^2}})~(\frac{\sqrt{4-u^2}}{\sqrt{4-u^2}})$
$sec~\theta = \frac{2~\sqrt{4-u^2}}{4-u^2}$
Therefore, $~~sec~(arccot~\frac{\sqrt{4-u^2}}{u}) = \frac{2~\sqrt{4-u^2}}{4-u^2}$
