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March 27, 2017

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Given that tan A = 2 tan B (B is the symbol for Beta) show that tan (A-B) = sin 2B/3- cos2 Beta.

  • trigonometry - ,

    tan(A-B) = (tanA - tanB)/(1+tanAtanB)
    but tanA = 2tanB
    so above
    = (2tanB - tanB)/(1 + 2tanBtanB)
    = tanB/(1+2tan^2B)
    = (sinB/cosB) / (1 + 2sin^2 B /cos^2 B)
    = sinB/cosB /[(cos^2 B + 2sin^2 B)/cos^2 B]

    = (sinB/cosB) * (cos^2 B)/(cos^2 B + 2sin^2 B)
    = sinBcosB /(cos^2 B + 2(1 - cos^2 B) )
    = sinBcosB / (2 - cos^2 B)
    = 2sinBcosB/ (4 - 2cos^2 B)
    = sin 2B / (4 - 2cos^2 B)

    which is different from yours,
    I can't seem to find my error if there is one.

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