Answer
$-\frac{48+25\sqrt 3}{39}$
Work Step by Step
Step 1. Letting $sin^{-1}(-\frac{1}{2})=x$, we have $sin(x)=-\frac{1}{2}, cos(x)=\frac{\sqrt 3}{2}$ and $tan(x)=-\frac{\sqrt 3}{3}$
Step 2. Letting $tan^{-1}\frac{3}{4}=y$, we have $tan(y)=\frac{3}{4}$
Step 3. We have $tan(x-y)=\frac{tan(x)-tan(y)}{1+tan(x)tan(y)}=\frac{-\frac{\sqrt 3}{3}-(\frac{3}{4})}{1+(-\frac{\sqrt 3}{3})(\frac{3}{4})}=-\frac{4\sqrt 3+9}{3(4-\sqrt 3)}\times\frac{4+\sqrt 3}{4+\sqrt 3}=-\frac{48+25\sqrt 3}{39}$
