tanC=tan(π-A-B)=-tan(A+B)=-(tanA+tanB)/(1-tanAtanB)
即:tanC=-(tanA+tanB)/(1-tanAtanB)=-tanAtanB/(1-tanAtanB)
令tanA=a,tanB=b,tanC=c
则c=-ab/(1-ab)=ab/(ab-1)=1+1/(ab-1)
因为tanA+tanB=tanAtanB,即:a+b=ab
又因为a+b≥2倍根号下ab,所以可得ab≥4
ab-1≥3,0<1/(ab-1)≤1/3
1<1+1/(ab-1)≤4/3
即c的最大值为4/3