已知2tanA=3tanB。求证:tan(A-B)=sin2B/5-cos2B
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tan(A-B)=(tanA-tanB)/(1+tanAtanB)
=(tanB/2)/(1+3tan^2B/2)
=tanB/(2+3tan^2B)
=sinBcosB/(2cos^2B+3sin^2B)
=2sinBcosB/(4+2sin^2B)
=sin2B/(4+1-cos2B)
=sin2B/(5-cos2B)
=(tanB/2)/(1+3tan^2B/2)
=tanB/(2+3tan^2B)
=sinBcosB/(2cos^2B+3sin^2B)
=2sinBcosB/(4+2sin^2B)
=sin2B/(4+1-cos2B)
=sin2B/(5-cos2B)
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