Answer
$$\frac{1}{{\sqrt {1 + {x^2}} }}$$
Work Step by Step
$$\eqalign{
& {\text{From the triangle bellow we have that}} \cr
& \cos \theta = \frac{{{\text{Adjacent side}}}}{{{\text{Hypotenuse}}}} \cr
& \cos \theta = \frac{1}{{\sqrt {1 + {x^2}} }} \cr
& and \cr
& {\text{tan}}\theta = \frac{{{\text{Opposite side}}}}{{{\text{Adjacent side}}}} \cr
& {\text{tan}}\theta = x\,\,\, \cr
& \theta = {\tan ^{ - 1}}x \cr
& \cr
& {\text{Then,}} \cr
& \cos \theta = \cos \left( {{{\tan }^{ - 1}}x} \right) \cr
& \cos \left( {{{\tan }^{ - 1}}x} \right) = \frac{1}{{\sqrt {1 + {x^2}} }} \cr} $$
