Precalculus (6th Edition) Blitzer

$\frac{33}{65}$
Step 1. Given $sin\alpha=\frac{3}{5}, \frac{\pi}{2}\lt\alpha\lt\pi$, we know $\alpha$ is in quadrant II; thus $cos\alpha=-\sqrt {1-sin^2\alpha}=-\frac{4}{5}$ and $tan\alpha=-\frac{3}{4}$ Step 2. Given $cos\beta=-\frac{12}{13}, \pi\lt\beta\lt \frac{3\pi}{2}$, we know $\beta$ is in quadrant III; thus $sin\beta=-\sqrt {1-cos^2\beta}=-\frac{5}{13}$ and $tan\beta=\frac{5}{12}$ Step 3. Using the Subtraction Formula, we have $cos(\alpha-\beta)=cos\alpha cos\beta + sin\alpha sin\beta =(-\frac{4}{5})(-\frac{12}{13})+(\frac{3}{5})(-\frac{5}{13})=\frac{33}{65}$