Answer
$\sin{(A+B)} = - \dfrac{16}{65}$
$\cos{(A+B)} = \dfrac{63}{65}$
$\tan{(A+B)} =- \dfrac{16}{63}$
$QIV$
Work Step by Step
$\cos{A} = - \sqrt{1-(\dfrac{3}{5})^2} = - \dfrac{4}{5}$
$\cos{B} = - \sqrt{1- (\dfrac{-5}{13})^2} = - \dfrac{12}{13}$
$\sin{(A+B)} = \sin{A} \cos{B} + \cos{A} \sin{B}$
$\sin{(A+B)} = \sin{A} \cos{B} + \sin{B} \cos{A} = \dfrac{3}{5} \times \dfrac{-12}{13}+ \dfrac{-5}{13} \times \dfrac{-4}{5}$
$\sin{(A+B)} = - \dfrac{16}{65}$
$\cos{(A+B)} = \cos{A} \cos{B} - \sin{A} \sin{B}= \dfrac{-4}{5} \times \dfrac{-12}{13}- \dfrac{3}{5}\times \dfrac{-5}{13}$
$\cos{(A+B)} = \dfrac{63}{65}$
$\tan{(A+B)} = \dfrac{\sin{(A+B)}}{\cos{(A+B)}} = - \dfrac{16}{63}$
$\sin{(A+B)} < 0 \hspace{20pt} \cos{(A+B)} > 0 \hspace{20pt} \therefore QIV$
