设三角形ABC的内角A B C的对狡辩分别是a b c 且A=60度 c=3b 求

a²=(c/3)²+c²-2(c/3)c*cos60º=10c²/9-c²/3=7c²/9
∴a/c=√7/3
a/sinA=c/sinC===>sinC=(c/a)sin60º=3(√3/2)/√7=3√21/14
sinB=(b/c)sinC=(1/3)sinC=√21/14
∴1/tanB+1/tanC=cosB/sinB+cosC/sinC=sin(B+C)/(sinBsinC)
=sinA/[(√21/14)(3√21/14)]=(√3/2)/[9/28]=14√3/9