若tanB+tanC-√3·tanB·tanC=-√3则tanB+tanC=-√3(1-tanB·tanC)即tan(B+C)=(tanB+tanC)/(1-tanB·tanC)=-√3从而　tanA=-tan(B+C)=√3代入第二个式子,得3√3+√3·tanB +1=√3·tanB3√3+1=0,不成立．...