证明A+B+C=nπ（n∈Z）的充要条件是tanA+tanB+tanC=tanA*tanB*tanC

充分条件:
A+B+C=nπ（n∈Z）
A=nπ-B-C
tanA+tanB+tanC=tan(nπ-B-C)+tanB+tanC
=tanB+tanC-tan(B+C)
=tanB+tanC-(tanB+tanC)/(1-tanB*tanC)
=(tanB+tanC)*[1-1/(1-tanB*tanC)]
=(tanB+tanC)*[(-tanB*tanC)/(1-tanB*tanC)]
=(tanB+tanC)/(1-tanB*tanC)*(-tanB*tanC)
=-tan(B+C)*tanB*tanC
=tanA*tanB*tanC
必要条件：
tanA+tanB+tanC=tanA*tanB*tanC
tanA+tanB+tanC=tanB+tanC-(tanB+tanC)/(1-tanB*tanC)
tanA+tanB+tanC=tanB+tanC-tan(B+C)
tanA=-tan(B+C)
A+B+C=nπ
即得A+B+C=nπ（n∈Z）的充要条件是tanA+tanB+tanC=tanA*tanB*tanC