求证:三角形中 ,tanAccotB+tanBcotC+tanCcotA=1

由A+B+C=π得
tanA=-tan(π-A)=-tan(B+C)
再利用和角公式得
tan(B+C)=(tanB+tanC)/(1-tanB*tanC)
代入上式
tanA=-(tanB+tanC)/(1-tanB*tanC)
tanA(tanB*tanC-1)=tanB+tanC
tanA*tanB*tanC=tanA+tanB+tanC
上式两边同除tanA*tanB*tanC得
1=tanA/(tanAtanB*tanC)+tanB/(tanAtanB*tanC)+tanC/(tanAtanB*tanC)
即1/(tanB*tanC)+1/(tanA*tanC)+1/(tanA*tanB)=1
cotA*cotB+cotBcotC+cotCcotA=1
与1楼答案一样,原题有误.1楼答案正确.

A=45,B=60,C=75,tanA=1,tanB=√3,tanC=2+√3
cotA=1,cotB=√3/3,cotC=2-√3
tanAcotB+tanBcotC+tanCcotA=1*(√3/3)+√3*(2-√3)+(2+√3)*1不等于1
cotA*cotB+cotBcotC+cotCcotA=1*(√3/3)+(√3/3)*(2-√3)+(2-√3)*1=1

tan(A)tan(B)tan(C) 不等于 0.

tan(C) = tan(PI - A - B) = -tan(A+B) = -[tan(A) + tan(B)]/[1 - tan(A)tan(B)],

tan(C)[1 - tan(A)tan(B)] = -[tan(A) + tan(B)],

tan(A) + tan(B) + tan(C) = tan(A)tan(B)tan(C),

cot(B)cot(C) + cot(A)cot(C) + cot(A)cot(B) = 1.