Answer
$-\frac{16}{63}$
Work Step by Step
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 Addition Formula, we have
$tan(\alpha+\beta)=\frac{tan\alpha +tan\beta }{1-tan\alpha tan\beta }=\frac{(-\frac{3}{4})+(\frac{5}{12})}{1-(-\frac{3}{4})(\frac{5}{12})}=-\frac{16}{63}$
