求证SIN（A加B加C）/COSACOSBCOSC=TANA加TANB加TANC-TANATANBTANC
热心网友
sin(A+B+C)=sin[(A+B)+C]=sin(A+B)cosC+cos(A+B)sisC=(sinAcosB+cosAsinB)cosC+(cosAcosB-sinAsinB)sinC=sinAcosBcosC+cosAsinBcosC+cosAcosBsinC-sinAsinBsinC等式的两边同时除以cosAcosBcosC得到sin(A+B+C)/(cosAcosBcosC)=sinA/cosA+sinB/cosB+sinC/cosC-sinA/cosA*sinB/cosB*sinC/cosC=tanA+tanB+tabC-tanAtanBtanC.