设三角形ABC的内角A,B,C的对边分别为a,b,c,且A等于60度，c=3b.求cotB+cotC

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cotB+cotC=cosB/sinB+cosC/sinC=(cosBsinC+sinBcosC)/(sinBsinC)=sin(B+C)/(sinBsinC)

=sin(180°-A)/(sinBsinC)=sinA/(sinBsinC)

又由正弦定理，a/sinA=b/sinB=c/sinC=2R,所以，

cotB+cotC=[a^2/(bc)]/sinA=[a^2/(3b^2)]/sin60°=[(2√3)/9]*a^2/b^2

再由余弦定理，a^2=b^2+c^2-2bccosA=b^2+9b^2-2b*3b*(1/2)=7b^2知a^2/b^2=7

所以，cotB+cotC=(14√3)/9.

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