Answer
$$ - \frac{{\sqrt {15} }}{7}$$
Work Step by Step
$$\eqalign{
& \tan \left( {2{{\cos }^{ - 1}}\frac{1}{4}} \right) \cr
& {\text{Let }}\theta = {\cos ^{ - 1}}\frac{1}{4}{\text{, thus}} \cr
& \cos \theta = \frac{1}{4} \cr
& {\text{Recall that }}\cos \theta = \frac{{{\text{adjacent side}}}}{{{\text{hypotenuse}}}} \cr
& {\text{adjacent side}} = 1 \cr
& {\text{hypotenuse}} = 4 \cr
& {\text{opposite side }} = \sqrt {15} \cr
& \tan \theta = \frac{{{\text{opposite side}}}}{{{\text{adjacent side}}}} = \sqrt {15} \cr
& \cr
& {\text{We have that }}\tan \left( {2{{\cos }^{ - 1}}\frac{1}{4}} \right) = \tan \left( {2\theta } \right) \cr
& {\text{Use the identity }}\tan 2\theta = \frac{{2\tan \theta }}{{1 - {{\tan }^2}\left( \theta \right)}} \cr
& \tan \left( {2{{\cos }^{ - 1}}\frac{1}{4}} \right) = \frac{{2\sqrt {15} }}{{1 - {{\left( {\sqrt {15} } \right)}^2}}} \cr
& \tan \left( {2{{\cos }^{ - 1}}\frac{1}{4}} \right) = - \frac{{\sqrt {15} }}{7} \cr} $$
