CBSE Class 11-science Answered
trigonometry
Asked by | 21 Sep, 2008, 09:29: PM
Expert Answer
let us denote α = A and β = B
2tanA = 3tanB ==> tanA =3/2 tanB
tan(A-B) = (tanA - tanB) / ( 1+tanAtanB) = 1/2 tanB / ( 1+3/2 tan2B)
put tanB = sinB / cosB and simplify
= sinBcosB / (2cos2B + 3sin2B ) = 2sinBcosB / (4cos2B + 6sin2B) = sin2B / (4 +2sin2B)
=sin2B / (4 - (1-2sin2B)+1)
=sin2B / (5 - cos2B)
Answered by | 21 Sep, 2008, 10:59: PM
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