Answer
$-\frac{48+25\sqrt 3}{39}$
Work Step by Step
1. Let $sin^{-1}(-\frac{1}{2})=u$ (u in quadrant IV), we have $sin(u)=-\frac{1}{2}$ and $tan(u)=-\frac{\sqrt 3}{3}$
2. Let $tan^{-1}\frac{3}{4}=v$, we have $tan(v)=\frac{3}{4}$
3. Thus $tan(u-v)=\frac{tan(u)-tan(v)}{1+tan(u)tan(v)}=\frac{-\frac{\sqrt 3}{3}-\frac{3}{4}}{1-\frac{\sqrt 3}{3}(\frac{3}{4})}=-\frac{4\sqrt 3+9}{12-3\sqrt 3}
=-\frac{4\sqrt 3+9}{12-3\sqrt 3}\times\frac{12+3\sqrt 3}{12+3\sqrt 3}
=-\frac{48+25\sqrt 3}{39}$
