Amaths

prove sin(A+B)/sin(A-B)=(tanA+tanB)/(tanA-tanB)

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  1. tan(A+B) = (tanA+tanB)/(1-tanAtanB)
    tan(A-B) = (tanA-tanB)/(1+tanAtanB)

    so,

    tanA+tanB = tan(A+B)(1-tanAtanB)
    tanA-tanB = tan(A-B)(1+tanAtanB)

    divide the two and you have

    tan(A+B)/tan(A-B) * (1-tanAtanB)/(1+tanAtanB)

    multiply top and bottom by cosAcosB nad you have

    tan(A+B)/tan(A-B) * (cosAcosB-sinAsinB)/(cosAcosB+sinAsinB)

    That's just

    tan(A+B)/tan(A-B) * cos(A+B)/cos(A-B)

    = sin(A+B)/sin(A-B)

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    posted by Steve
  2. very good

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    posted by timothy

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