trigonometry26

If 1+sin^2 theta = 3sin theta cos theta, then prove that tan theta =1or 1/2

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  1. 1 + sin^2 Ø = 3sinØcosØ
    divide by cosØ
    1/cosØ + tanØsinØ = 3sinØ
    secØ + tanØsinØ = 3sinØ
    tanØ = (3sinØ - secØ)/sinØ
    = 3 - secØ/sinØ
    let's stop here:
    tanØ = 3 - 1/(sinØcosØ) ******

    if tanØ = 1 , sinØ/cosØ = 1 and sinØ = cosØ, Ø = π/2
    in *****
    LS = tanØ = 1
    RS = 3 - 1/sin^2 Ø
    = 3 - 1/(1/2) = 3-2 = 1
    = LS

    if tanØ = 1/2
    sinØ = 1/√5 and cosØ = 2/√5

    in *****
    LS = 1/2
    RS = 3 - 1/(sinØcosØ)
    = 3 - 1/((1/√5)(2/√5))
    = 3 - 1/(2/5)
    = 3 - 5/2
    = 1/2
    = LS

    graphic confirmation:
    http://www.wolframalpha.com/input/?i=solve+tanx+%3D+3+-+1%2F%28sinxcosx%29

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  2. or, you can try it like this

    1 + sin^2 Ø = 3sinØcosØ
    sin^2 Ø = (1-cos2Ø)/2

    1 + (1-cos2Ø)/2 = 3/2 sin2Ø
    3 - cos2Ø = 3sin2Ø
    9 - 6cos2Ø + cos^2 2Ø = 9 - 9cos^2 2Ø
    10cos^2 2Ø + 6cos2Ø = 0
    2cos2Ø(5cos2Ø-3) = 0

    cos2Ø = 0 or cos2Ø = 3/5

    so Ø = π/4 as above and tanØ=1, or

    tanØ = √((1-cos2Ø)/(1cos2Ø))
    = √(1 - 3/5)/(1 + 3/5) = √(1/4) = 1/2

    Those are the angles in the 1st quadrant that satisfy the equation.

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