Answer
$\frac{2\sqrt {65}-6\sqrt {13}}{39}$
Work Step by Step
Step 1. Letting $sin^{-1}\frac{2}{3}=x, 0\lt x\lt\frac{\pi}{2}$, we have $sin(x)=\frac{2}{3}, cos(x)=\frac{\sqrt {9-4}}{3}=\frac{\sqrt {5}}{3}$
Step 2. Letting $tan^{-1}\frac{3}{2}=y, 0\lt y\lt\frac{\pi}{2}$, we have $tan(y)=\frac{3}{2}, sin(y)=\frac{3}{\sqrt {9+4}}=\frac{3\sqrt {13}}{13},cos(y)=\frac{2\sqrt {13}}{13}$
Step 3. Thus $cos(x+y)=cos(x)cos(y)-sin(x)sin(y)=(\frac{\sqrt {5}}{3})(\frac{2\sqrt {13}}{13})-(\frac{2}{3})(\frac{3\sqrt {13}}{13})=\frac{2\sqrt {65}-6\sqrt {13}}{39}$
