tanA=tan[π-(B+C)]=-tan(B+C)=-(tanB+tanC)/(1-tanB*tanC),
——》tanA(1-tanB*tanC)=-(tanB+tanC),
——》tanA+tanB+tanc=tanAtanBtanC,
命题得证.
tanA=tan[π-(B+C)]=-tan(B+C)=-(tanB+tanC)/(1-tanB*tanC),
——》tanA(1-tanB*tanC)=-(tanB+tanC),
——》tanA+tanB+tanc=tanAtanBtanC,
命题得证.