在三角形中,有cosA=-cos(B+C）,则所求式=-cos(B+C)/cos(B-C),=-cos(B+C）=sinBsinC-cosBcosC,cos(B-C)=cosBcosC+sinBsinC,又5tanBtanC=1,则sinBsinC/cosBcosC=1/5,则sinBsinC-cosBcosC/cosBcosC+sinBsinC=-4/6=-2/3....