已知2tanA=3tanB,求证tan(A-B)=sin2B/5-cos2B
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已知2tanA=3tanB,求证tan(A-B)=sin2B/5-cos2B已知2tanA=3tanB,求证tan(A-B)=sin2B/5-cos2B已知2tanA=3tanB,求证tan(A-B)
已知2tanA=3tanB,求证tan(A-B)=sin2B/5-cos2B
已知2tanA=3tanB,求证tan(A-B)=sin2B/5-cos2B
已知2tanA=3tanB,求证tan(A-B)=sin2B/5-cos2B
如图：(点开并最大化……)
tan(A-B)=(tanA-tanB)/(1+tanA*tanB)=tanB/2(1+3/2*tanB^2)=tanB/(2+3tanB^2)
sin2B/(5-cos2B)=2sinBcosB/(6-2cosB^2)=sinBcosB/(3-cosB^2)=sinB/cosB/(3/cosB^2-1)=tanB/(3/cosB^2-1)=tanB/(2+3tanB^2)
所以 tan(A-B)=sin2B/5-cos2B
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