首先证明这样一个结论:三角形ABC tanAtanBtanC=tanA+tanB+tanC证明如下tanA=tan(∏-B-C)=-tan(B+C)= -(tanB+tanC)/(1-tanBtanC)=（tanB+tanC)/(tanBtanC-1)所以 tanA*(tanBtanC-1)=tanB+tanCtanA*tanB*tanC - tanA=t...