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

2个回答

  • 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