已知tan(A-B)=sin2B/(5-cos2B) 求证:2tanA=3tanB

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已知tan(A-B)=sin2B/(5-cos2B) 求证:2tanA=3tanB
已知tan(A-B)=sin2B/(5-cos2B) 求证:2tanA=3tanB

已知tan(A-B)=sin2B/(5-cos2B) 求证:2tanA=3tanB
证明：
tan(A-B)=sin2B/(5-cos2B)
左边=(tanA-tanB)/(1+tanAtanB)
右边=sin2B/(5-cos2B)
=(2sinBcosB)/(6-2cos²B)
=(2sinBcosB)/(6sin²B+4cos²B)
=(sinBcosB)/(2cos²B+3sin²B)
=tanB/(2+3tan²B)
即 (tanA-tanB)/(1+tanAtanB) =tanB/(2+3tan²B)
∴ (tanA-tanB)*(2+3tan²B)=tanB(1+tanAtanB)
即 2tanA+3tanAtan²B-2tanB-3tan³B=tanB+tanAtan²B
∴ 2tanA-3tanB+2tanAtan²B-3tan³B=0
即 (2tanA-3tanB)+tan²B(2tanA-3tanB)=0
即 (2tanA-3tanB)(1+tan²B)=0
∵ 1+tan²B恒不为0
则2tanA-3tanB=0
即 2tanaA=3tanB

因为tan(A-B)=sin2B/(5-cos2B),
展开(tanA-tanB)/(1+tanAtanB)=[2tanB/(1+tan^B)]/[5-(1-tan^B)/(1+tan^B)],其中tan^B=(tanB)^2,
化简右边得(tanA-tanB)/(1+tanAtanB)=2tanB/(4+6tan^B),
去分母得（tanA-tanB)(2+3tan^B)=tanB(1+tanAtanB),
整理得tanA(2+2tan^B)=tanB(3+3tan^B),
1+tan^B>0,
∴2tanA=3tanB.

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