Answer
$$\frac{{{\text{2}}\sqrt {4 - {u^2}} }}{{4 - {u^2}}}$$
Work Step by Step
$$\eqalign{
& \sec \left( {\operatorname{arccot} \frac{{\sqrt {4 - {u^2}} }}{u}} \right) \cr
& {\text{From the triangle we have that}} \cr
& \cot \theta = \frac{{\sqrt {4 - {u^2}} }}{u} \cr
& \theta = \operatorname{arccot} \frac{{\sqrt {4 - {u^2}} }}{u} \cr
& \sec \left( {\operatorname{arccot} \frac{{\sqrt {4 - {u^2}} }}{u}} \right) = \sec \theta \cr
& \sec \left( {\operatorname{arccot} \frac{{\sqrt {4 - {u^2}} }}{u}} \right) = \frac{{{\text{hypotenuse}}}}{{{\text{adjacent side}}}} \cr
& \sec \left( {\operatorname{arccot} \frac{{\sqrt {4 - {u^2}} }}{u}} \right) = \frac{{\text{2}}}{{\sqrt {4 - {u^2}} }} \cr
& {\text{Rationalizing}} \cr
& \sec \left( {\operatorname{arccot} \frac{{\sqrt {4 - {u^2}} }}{u}} \right) = \frac{{\text{2}}}{{\sqrt {4 - {u^2}} }} \times \frac{{\sqrt {4 - {u^2}} }}{{\sqrt {4 - {u^2}} }} \cr
& \sec \left( {\operatorname{arccot} \frac{{\sqrt {4 - {u^2}} }}{u}} \right) = \frac{{{\text{2}}\sqrt {4 - {u^2}} }}{{4 - {u^2}}} \cr} $$
