Asked by Tay
what is tan(A-B) if SinA=1/2, cosA>0, tanB=3/4 and sinB<0
Answers
Answered by
Steve
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
tan(A-B) = (tanA-tanB)/(1+tanAtanB)
= (1/√3 - 3/4)/(1+(1/√3)(3/4))
= (25√3 - 48)/39
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