If $\displaystyle {\sin{\alpha}}=-\frac{{2}}{{3}}$ such that $\displaystyle \alpha$ terminates in quadrant III and $\displaystyle {\cos{\beta}}=-\frac{{3}}{{4}}$ such that $\displaystyle \beta$ terminates in quadrant II then

 $\displaystyle {\tan{{\left(\alpha+\beta\right)}}}=$

Your answer should be exact but you do not have to rationalize your denominator.