1)1/tanA+1/tanC=(cosA*sinC+cosC*sinA)/(sinA*sinC)=sin(A+C)/(sinA*sinC)=sinB/(sinA*sinC)
由题意得sinA,sinB,sinC也成等比数列,所以sinB的平方等于sinA*sinC=(5/13)*(5/13)=25/169
代入的结果为1/sinB=13/5
2)ac=b^2,cosB=12/13,b^2=ac=13
由余弦定理b^2=a^2+c^2-2accosB 推出,a^2+c^2=37
(a+c)^2=a^2+c^2+2ac=37+26=63
a+c=3√7