If $\displaystyle {\csc{\alpha}}=-\frac{{9}}{{8}}$ such that $\displaystyle \alpha$ terminates in quadrant IV and $\displaystyle {\sec{\beta}}=-{3}$ such that $\displaystyle \beta$ terminates in quadrant II then

 $\displaystyle {\sin{{\left(\alpha+\beta\right)}}}=$Preview Question 6  

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