Answer
$\frac{8\sqrt 2-9\sqrt 3}{5}$
Work Step by Step
Step 1. Using the given figures, we have $x=\sqrt {4-1}=\sqrt 3$ and $y=\sqrt {1-1/9}=-\frac{2\sqrt 2}{3}$
Step 2. We have $sin(\alpha)=\frac{1}{2}, cos(\alpha)=\frac{\sqrt 3}{2}, tan\alpha=\frac{\sqrt 3}{3}, sin(\beta)=-\frac{2\sqrt 2}{3}, cos(\beta)=\frac{1}{3}, tan\beta=-2\sqrt 2$
Step 3. $h(\alpha+\beta)=tan(\alpha+\beta)=\frac{tan(\alpha)+tan(\beta)}{1-tan(\alpha)tan(\beta)}=\frac{(\frac{\sqrt 3}{3})+(-2\sqrt 2)}{1-(\frac{\sqrt 3}{3})(-2\sqrt 2)}=\frac{\sqrt 3-6\sqrt 2}{3+2\sqrt 6}\times\frac{3-2\sqrt 6}{3-2\sqrt 6}=\frac{3\sqrt 3-6\sqrt 2-18\sqrt 2+24\sqrt 3}{9-24}=\frac{8\sqrt 2-9\sqrt 3}{5}$
