Answer
$ \tan\left ( \sin ^{-1}\frac{u}{\sqrt{u^2+2}} \right )=\frac{u\sqrt 2}{2} $
Work Step by Step
Let $ \theta = \sin ^{-1}\frac{u}{\sqrt{u^2+2}} $
then $ \sin\theta = \frac{u}{\sqrt{u^2+2}} $
By using trigonometric identities.
$ \cos^2\theta+\sin^2\theta=1 $
Isolate $ \cos \theta $.
$ \cos \theta = \sqrt{1-\sin^2 \theta} $
Substitute the value of $ \sin \theta $ .
$ \cos \theta = \sqrt{1-\left (\frac{u}{\sqrt{u^2+2}} \right )^2} $
Clear parentheses.
$ \cos \theta = \sqrt{1-\frac{u^2}{u^2+2} } $
Simplify.
$ \cos \theta = \sqrt{\frac{u^2+2-u^2}{u^2+2} }
$
$\cos \theta = \sqrt{\frac{2}{u^2+2} }
$
$\cos \theta = \frac{\sqrt{2}}{\sqrt{u^2+2}}
$
By using trigonometric ratios
$\tan \theta = \frac{\sin \theta }{\cos \theta } $
Substitute the values of $ \sin \theta $ and $\cos \theta $.
$ \tan \theta = \frac{\frac{u}{\sqrt{u^2+2}} }{\frac{\sqrt{2}}{\sqrt{u^2+2}} } $
Simplify.
$\tan \theta = \frac{u}{\sqrt{2}}=\frac{u\sqrt 2}{2} $