Answer
$tan~(arcsec\frac{\sqrt{u^2+1}}{u}) = \frac{1}{u}$
Work Step by Step
$\theta = arcsec(\frac{\sqrt{u^2+1}}{u})$
$sec~\theta = \frac{\sqrt{u^2+1}}{u} = \frac{hypotenuse}{adjacent}$
Note that $\theta$ is in quadrant I since $u \gt 0$. We can find the magnitude of the opposite side:
$opposite = \sqrt{(\sqrt{u^2+1})^2-u^2} = 1$
We can find the value of $tan~\theta$:
$tan~\theta = \frac{opposite}{adjacent}$
$tan~\theta = \frac{1}{u}$
Therefore, $tan~(arcsec\frac{\sqrt{u^2+1}}{u}) = \frac{1}{u}$
