# Trigonometry

Sin A =1/2 and Cos B=12/13 where A lies in 1 quadrant and B lies in 3 quadrant. Then find tan(a-b).

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1. error in data
if B is in quad III, then cosB must be negative
I will assume you meant:
cos B = -12/13

make sketches of triangles in the two quadrants
for sinA = 1/2, y = 1, r = 2
x^2 + 1^2 = 2^2
x = √3
so tanA = y/x = 1/√3

for cosB = -12/13 in III
x = -12 , r = 13
then in III, y = -5
tanB = -5/-12 = 5/12

tan(A-B) = (tanA - tanB)/( 1 + tanAtanB)
= ((1/√3 - 5/12)/(1 + (1/√3)(5/12)
= ( (12 - 5√3)/(12√3) ) / ( (12√3 + 5)/(12√3) )
= (12 - 5√3)/(12√3 + 5)

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