1.在锐角三角形中,证tanA+tanB+tanC=tanAtanBtanC

1个回答

  • tanA+tanB+tanC

    =tan(A+B)(1-tanAtanB)+tanC

    =tan(π-c)(1-tanAtanB)+tanC

    =-tanC(1-tanAtanB)+tanC

    =-tanC+tanAtanBtanC+tanC

    =tanAtanBtanC

    2.题目有误

    (sin9°+cos15°sin6°)/(cos9°-sin15°sin6°)

    =(sin(15-6)+cos15*sin6)/(cos(15-6)-sin15*sin6)

    =(sin15*cos6-cos15*sin6+cos15*sin6)/(cos15*cos6+sin15*sin6- sin15*sin6)

    =sin15*cos6/(cos15*cos6)

    =tan15 [tan(A/2)=(1-cosA)/sinA]

    =(1-cos30)/sin30

    =(1-√3/2)/(1/2)

    =2*(1-√3/2)

    =2-√3