Answer
$-\frac{8\sqrt 2+9\sqrt 3}{5}$
Work Step by Step
1. $h(\alpha-\beta)=tan(\alpha-\beta)=\frac{tan\alpha-tan\beta}{1+tan\alpha tan\beta} $
2. From the given figures, we have $tan\alpha=\frac{\sqrt 3}{3}, tan\beta=-2\sqrt 2$
3. Thus $h(\alpha-\beta)=\frac{\frac{\sqrt 3}{3}+2\sqrt 2}{1-\frac{\sqrt 3}{3}(2\sqrt 2)}=\frac{\sqrt 3+6\sqrt 2}{3-2\sqrt 6}=\frac{\sqrt 3+6\sqrt 2}{3-2\sqrt 6}\times\frac{3+2\sqrt 6}{3+2\sqrt 6}
=\frac{3\sqrt 3+6\sqrt 2+18\sqrt 2+24\sqrt 3}{-15}=\frac{27\sqrt 3+24\sqrt 2}{-15}=-\frac{8\sqrt 2+9\sqrt 3}{5}$
