calculus

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given 7sin^2 theta + cos theta sin theta = 6 show that tan^2 theta + tan theta - 6 = 0

  • calculus -

    I worked it backwards, since the second equation was easier to solve

    tan^2Ø + tanØ-6=0
    (tanØ-2)(tanØ+3) = 0
    tanØ = 2 or tanØ=-3

    case1: tanØ=2, Ø could be in I or III
    In I, sinØ = 2/√5 and cosØ = 1/√5
    test in first equation
    LS = 7(4/5) + (1/√5)(2/√5)
    = 28/5 + 2/5 = 30/5 = 6 = RS

    In III
    sinØ = -2/√5 , cosØ = -1/√5
    LS = 7(4/5) + (-2/√5)(-1/√5) = 30/5 = 6 = RS

    Case 2: tanØ = -3, Ø could be in II or IV
    in II , sinØ = 3/√10 , cosØ = -1/√10
    testing in first equation,
    LS = 7(9/10) + (-1/√10)((3/√10)
    = 63/10 - 3/10 = 60/10 = 6 = RS

    in IV, sinØ = -3/√10 , cosØ = 1/√10
    LS = 7(9/10) + (1/√10)(-3/√10)
    = 63/10 - 3/10 = 60/10 = 6 = RS

    so all 4 values of Ø satisfy both equations.

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