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ifm sin B=nsin(2A+B) prove that(m+n)tanA=(m-n)tan(A+B)

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  1. hint:

    (m-n)tan(A+B)
    = (m-n) (tanA+tanB)/(1-tanAtanB)
    = (mtanA - ntanB)/(1-tanAtanB)
    now multiply top and bottom by cosAcosB:
    = (msinAcosB-nsinBcosA)/(cosAcosB-sinAsinB)

    the rest should follow without difficulty.

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  2. Sum

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  3. What's this answer
    It's really unhelpful
    Do they really know the answer or they kept something for simply expecting ( like) from us

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