cscA = 5/3, where 90deg < A < 180deg

csc = hyp /opp

So A is in the 2nd quadrant.

The reference triangle of A has:

opposite side = 3

adjacent side = -4 ----------you know how to get this?

hypotenuse = 5

hence,

tanA = opp/adj = 3 /(-4) = -3/4

---

cosB = 5/13, where -90deg < B < 0deg

cos = adj /hyp

So, B is the 4th quadrant.

The reference right triangle of B has:

opp = -12 -----how about this?

adj = 5

hyp = 13

hence,

tanB = -12/5

Then,

tan(A +B)

= [tanA +tanB] / [ 1 -tanAtanB]

= [-3/4 -12/5] / [ 1 -(-3/4)(-12/5)]

= 63/16 ------------------------------answer.