Question

在△ABC中,若tanA（tanB-tanC）=tanBtanC,则（sinA/sinC）^2+（sinB/sinC）^

Answer 1

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