题目内容：

在三角形ABC中,A,B,C的对边分别为a,b,c,已知sinB=5/13,且a,b,c成等比数列,求cotA+cotC的值

优质解答

a,b,c成等比数列,则b^2=ac
由正弦定理得sinAsinC=(sinB)^2
cotA+cotC=cosA/sinA+cosC/sinC
=(cosAsinC+cosCsinA)/(sinAsinC)
=sin(A+C)/(sinAsinC)
=sinB/(sinB)^2
=1/sinB
=1/(5/13)
=13/5