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

Answer 1

make quick sketches of right-angled triangles for each of the two cases, and use Pythagoras to find the missing sides.

first triangle:
since sinA = 12/13 , y = 12 and r = 13
x^2 + y^2 = r^2
x^2 + 144 = 169
x^2 = 25
x = ±5, but we are in quad II, so x = -5
then tanA = - 12/5

2nd triangle:
since tanB = -4/3

tan(A-B) = (tanA - tanB)/(1 + tanAtanB)
= (-12/5 - (-4/3) )/( 1 + (-12/5)(-4/3)
= (-16/15) / (21/5)
= -16/63

Answer 2

Are you sure it's negative?