Answer
$$\frac{{u\sqrt 2 }}{2}$$
Work Step by Step
$$\eqalign{
& \tan \left( {{{\sin }^{ - 1}}\frac{u}{{\sqrt {{u^2} + 2} }}} \right) \cr
& {\text{From the triangle we have that}} \cr
& \sin \theta = \frac{u}{{\sqrt {{u^2} + 2} }} \cr
& \theta = {\sin ^{ - 1}}\frac{u}{{\sqrt {{u^2} + 2} }} \cr
& \tan \left( {{{\sin }^{ - 1}}\frac{u}{{\sqrt {{u^2} + 2} }}} \right) = \tan \theta \cr
& \tan \left( {{{\sin }^{ - 1}}\frac{u}{{\sqrt {{u^2} + 2} }}} \right) = \frac{{{\text{opposite side}}}}{{{\text{adjacent side}}}} \cr
& \tan \left( {{{\sin }^{ - 1}}\frac{u}{{\sqrt {{u^2} + 2} }}} \right) = \frac{u}{{\sqrt 2 }} = \frac{{u\sqrt 2 }}{2} \cr} $$
