在斜三角形ABC中,tanA+tanB+tanC-tanAtanBtanC的值是多少?
问题描述：
在斜三角形ABC中,tanA+tanB+tanC-tanAtanBtanC的值是多少?
答
tanA+tanB+tanC-tanAtanBtanC
=sinA/cosA+sinB/cosB+sinC/cosC-sinAsinBsinC/cosAcosBcosC
=(sinAcosBcosC+sinBcosAcocC+sinCcosAcosB-sinAsinBsinC)/cosAcosBcosC
=[sinA(cosBcosC-sinBsinC)+cosA(sinBcosC+sinCcosB)]/cosAcosBcosC
=[sinAcos(B+C)+cosAsin(B+C)]/cosAcosBcosC
=sin(A+B+C)/cosAcosBcosC
=sin180度/cosAcosBcosC
sin180度=0
所以原式=0