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

make quick sketches of rightangled 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(AB) = (tanA  tanB)/(1 + tanAtanB)
= (12/5  (4/3) )/( 1 + (12/5)(4/3)
= (16/15) / (21/5)
= 16/63 
Are you sure it's negative?