Question

在三角形ABC中，证明tanA+tanB+tanC=tanA*tanB*tanC

Answer 1

∵A+B＝π-C，
∴tan(A+B)=tan(π-C)
即：(tanA+tanB)/(1-tanA*tanB)=-tanC,
∴tanA+tanB=-tanC(1-tanAtanB)
即：tanA+tanB=-tanC+tanAtanBtanC
∴tanA+tanB+tanC=tanAtanBtanC

Answer 2

∵A+B＝π-C，
∴tan(A+B)=tan(π-C)
(tanA+tanB)/(1-tanA*tanB)=-tanC,
tanA+tanB=-tanC+tanAtanBtanC
∴tanA+tanB+tanC=tanAtanBtanC
所以tanAtanBtanC=(tanA+tanB+tanC)

Answer 3

tan(B+C) = (tanB+tanC)/(1-tanBtanC)
tanB+tanC=(1-tanBtanC)tan(B+C)
=-(1-tanBtanC) tanA
tanA+tanB+tanC
=tanA-(1-tanBtanC) tanA
=tanAtanBtanC