Posted by **Anonymous** on Tuesday, September 17, 2013 at 11:22pm.

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

- Trig -
**Reiny**, Wednesday, September 18, 2013 at 12:22am
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

- Trig -
**TeeKay**, Wednesday, October 15, 2014 at 2:45pm
Are you sure it's negative?

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