Answer
$$\frac{1}{{\sqrt 5 }}$$
Work Step by Step
$$\eqalign{
& \cos \left( {{{\tan }^{ - 1}}\left( { - 2} \right)} \right) \cr
& = \cos \left( {{{\tan }^{ - 1}}\left( 2 \right)} \right) \cr
& {\text{Let }}\theta = {\tan ^{ - 1}}\left( 2 \right){\text{, thus}} \cr
& \tan \theta = 2 \cr
& {\text{Recall that }}\tan \theta = \frac{{{\text{opposite side}}}}{{{\text{adjacent side}}}} \cr
& {\text{opposite side}} = 2 \cr
& {\text{adjacent side}} = 1 \cr
& {\text{hypotenuse}} = \sqrt 5 \cr
& {\text{Then}} \cr
& \cos \theta = \cos \left( {{{\tan }^{ - 1}}\left( 2 \right)} \right) = \frac{{{\text{adjacent side}}}}{{{\text{hypotenuse}}}} \cr
& \cos \left( {{{\tan }^{ - 1}}\left( { - 2} \right)} \right) = \frac{1}{{\sqrt 5 }} \cr} $$
