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