若A+B+C=180°,证明tanA+tanB+tanC=tanAtanBtanC.
问题描述:
若A+B+C=180°,证明tanA+tanB+tanC=tanAtanBtanC.
答
C = 180° - A - B
所以tanC = tan(180° - A - B)= -tan(A + B)
左边 = tanA + tanB - tan(A + B)
= tanA + tanB - (tanA + tanB)/(1 - tanAtanB)
= [(tanA + tanB)(1 - tanAtanB)- (tanA + tanB)]/(1 - tanAtanB)
= -(tanA + tanB)(tanAtanB)/(1 - tanAtanB)
= tanAtanB [-(tanA + tanB)/(1 - tanAtanB)]
= tanAtanBtanC
= 右边
所以得证.