应该求tanB+tanC吧!由sinA=sin[∏-(B+C)]=即sin(B+C)=2cosBcosC,展开得,sinBcosC+sinCcosB=2cosBcosC,sinBcosC-cosBcosC=cosBcosC-sinCcosBcosC(sinB-cosB)=cosB(cosC-sinC)即(sinB-cosB)/cosB=(cosC-sinC)/cosCtanB...