Answer
$$\frac{{\sqrt {{u^2} + 5} }}{u}$$
Work Step by Step
$$\eqalign{
& \sec \left( {{{\cos }^{ - 1}}\frac{u}{{\sqrt {{u^2} + 5} }}} \right) \cr
& {\text{From the triangle we have that}} \cr
& \cos \theta = \frac{u}{{\sqrt {{u^2} + 5} }} \cr
& \theta = {\cos ^{ - 1}}\frac{u}{{\sqrt {{u^2} + 5} }} \cr
& \sec \left( {{{\cos }^{ - 1}}\frac{u}{{\sqrt {{u^2} + 5} }}} \right) = \sec \theta \cr
& \sec \left( {{{\cos }^{ - 1}}\frac{u}{{\sqrt {{u^2} + 5} }}} \right) = \frac{{{\text{hypotenuse}}}}{{{\text{adjacent side}}}} \cr
& \sec \left( {{{\cos }^{ - 1}}\frac{u}{{\sqrt {{u^2} + 5} }}} \right) = \frac{{\sqrt {{u^2} + 5} }}{u} \cr} $$
