∵在△ABC中,A+B+C=π∴A=π-(B+C)∴sinA=sin[π-(B+C)]=sin(B+C)=sinBcosC+cosCsinB∵sinA=cosBcosC∴sinBcosC+cosBsinC=cosBcosC∴(sinBcosC)/(cosBcosC)+(cosBsinC)/(cosBcosC)=1∴sinB/cosB+sinC/cosC=1∴tanB+t...