Let sin A = 12/13 with 90º≤A≤180º and tan B = -4/3 with 270º≤B≤360º. Find tan (A - B).

Answer 1

sketch right-angled triangles to match

sinA = 12/13
and
tanB = -4/3

from the first , y=12, r=13, so x = -5
and tanA = -12/5

tan(A-b) = (tanA - tanB)/(1 + tanAtanB)
= (-12/5 + 4/3)/(1 + (-12/5)(-4/3))
= -16/63

Answer 2

4/3