trig

1. if sin 2A = Psin2B,
prove that tan(A+B)/tan(A-B)=P+1/P-1.

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  1. I have assumed the RS is (P+1)/(P-1) instead of the way it was typed.

    I have been working on this off and on since it was first posted yesterday and seem to be getting nowhere.
    from sin 2A = Psin 2B
    2sinAcosA = 2PsinBcosB
    P = sinAcosA/(sinBcosB)

    I then went to the right side of the identity
    RS = (P+1)/(P-1)
    = [(sinAcosA + sinBcosB)/(sinBcosB)] / [(sinAcosA - sinBcosB)/(sinBcosB)
    = (sinAcosA + sinBcosB)/(sinAcosA - sinBcosB)

    Now, if this truly is an identity, it should be equal to
    tan(A+B)/tan(A-B) for all values of A and B, which would make the tangent defined.

    So I let A = 40° and B = 10°
    LS = tan50/tan30 = appr. 2.06
    RS = (sin50cos50 + sin10cos10)/(sin50cos50 - sin10cos10) = appr. 0.321

    LS ≠ RS, so it is not an identity.

    Other thoughts are appreciated.

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