因为 tanA（tanB-tanC）=tanBtanC
即 sinA/cosA(sinB/cosB-sinC/cosC)=sinBsinC/cosBcosC
sinA(sinBcosC-cosBsinC)=cosAsinBsinC
sinAsinBcosC=sinC(sinAcosB+cosAsinB)=sinCsin(A+B)=(sinC)^2
（sinA/sinC）^2+（sinB/sinC）^2
=[ (sinA)^2+(sinB)^2]/(sinC)^2
=[ (sinA)^2+(sinB)^2]/sinAsinBcosC
=sinA/sinBcosC+sinB/sinAcosC
=sin(B+C)/sinBcosC+sin(A+C)/sinAcosC
= (sinBcosC+cosBsinC)/sinBcosC+(sinAcosC+cosAsinC)/sinAcosC
=2+(sinAcosBsinC+cosAsinBsinC)/sinAsinBcosC
=2+[sinC(sinAcosB+cosAsinB)/sinAsinBcosC
=2+sinCsin(A+B)/sinAsinBcosC
=2+(sinC)^2/sinAsinBcosC [将sinAsinBcosC=(sinC)^2代入原式]
=3