Determining in which quadrant an angle terminates. Let $A$ be in the fourth quadrant and let $\sec(A) = \dfrac{13}{5}$, Let $B$ be in the third quadrant and let $\csc(B)=\dfrac{-5}{3}$. Find $\sin(A + B)$ and determine in which quadrant $A + B$ terminates.

$\sec(A) = \dfrac{13}{5} \Rightarrow \cos(A) = \dfrac{5}{13}$

$\csc(B) = \dfrac{-5}{3} \Rightarrow \sin(B) = \dfrac{-3}{5}$

$\sin(A) = \dfrac{-12}{13}$ and $\cos(B) = \dfrac{-4}{5}$

$\sin(A + B) = \sin(A) \cos(B) + \sin(B) \cos(A)$ 

after performing the substitutions, I arrive at $\sin(A + B) = \dfrac{33}{65}$


$\cos(A + B) = \cos(A)\cos(B) - \sin(A)\sin(B)$

= $\dfrac{5}{13}(\dfrac{-4}{5}) - \dfrac{-12}{13}(\dfrac{-3}{5}) = \dfrac{-56}{65}$

Hint: Your calculations above are correct (I cleaned them up a little bit).  If sine is positive but cosine is negative, what quadrant are you in?  Does this give a hint to $A + B$?
