Answer
$$\frac{{120}}{{169}}$$
Work Step by Step
$$\eqalign{
& \sin \left( {2{{\tan }^{ - 1}}\frac{{12}}{5}} \right) \cr
& {\text{Let }}\theta = {\tan ^{ - 1}}\frac{{12}}{5}{\text{, thus}} \cr
& \tan \theta = \frac{{12}}{5} \cr
& {\text{Recall that }}\tan \theta = \frac{{{\text{opposite side}}}}{{{\text{adjacent side}}}} \cr
& {\text{opposite side }} = 12 \cr
& {\text{adjacent side}} = 5 \cr
& {\text{hypotenuse}} = 13 \cr
& \cr
& {\text{We have that }}\sin \left( {2{{\tan }^{ - 1}}\frac{{12}}{5}} \right) = \sin \left( {2\theta } \right) \cr
& {\text{Use the identity }}\sin 2\theta = 2\sin \theta \cos \theta \cr
& \sin \left( {2{{\tan }^{ - 1}}\frac{{12}}{5}} \right) = 2\sin \theta \cos \theta \cr
& \sin \left( {2{{\tan }^{ - 1}}\frac{{12}}{5}} \right) = 2\left( {\frac{{{\text{opposite side}}}}{{{\text{hypotenuse}}}}} \right)\left( {\frac{{{\text{adjacent side}}}}{{{\text{hypotenuse}}}}} \right) \cr
& \sin \left( {2{{\tan }^{ - 1}}\frac{{12}}{5}} \right) = 2\left( {\frac{{{\text{12}}}}{{{\text{13}}}}} \right)\left( {\frac{{\text{5}}}{{{\text{13}}}}} \right) \cr
& \sin \left( {2{{\tan }^{ - 1}}\frac{{12}}{5}} \right) = \frac{{120}}{{169}} \cr} $$
