Answer
$\frac{16}{65}$
Work Step by Step
1. Let $tan^{-1}(\frac{4}{3})=u$ (quadrant I), we have $tan(u)=\frac{4}{3}, sin(u)=\frac{4}{5}, cos(u)=\frac{3}{5}$
2. Let $cos^{-1}(\frac{12}{13})=v$ (quadrant I), we have $cos(v)=\frac{12}{13}, sin(v)=\frac{5}{13}, $
3. $cos(u+v)=cos(u)cos(v)-sin(u)sin(v)=(\frac{3}{5})(\frac{12}{13})-(\frac{4}{5})(\frac{5}{13})=\frac{16}{65}$
