what is tan(A-B) if SinA=1/2, cosA>0, tanB=3/4 and sinB<0

Answer 1

sinA = 1/2, so tanA = 1/√3

tan(A-B) = (tanA-tanB)/(1+tanAtanB)
= (1/√3 - 3/4)/(1+(1/√3)(3/4))
= (25√3 - 48)/39