在△ABC中,内角A,B,C的对边分别为a,b,c,已知a,b,c成等比数列,且cosB=3/4.求

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ac=b^2
即(sinA*2R)(sinC*2R)=(sinB*2R)^2
∴sinAsinC=(sinB)^2
cosA/sinA+cosC/sinC
=(cosAsinC+cosCsinA)/sinAsinC
=sin(A+C)/sinAsinC
=sin(π-B)/(sinB)^2
=1/sinB
（sinB>0)=1/√(1-cos^2B)
=1/√(1-9/16)
=4/√7=4√7/7
======以下答案可供参考======
供参考答案1:
b^2=ac，(sinB)^2=sinAsinC。
sinB=√[1-(cosB)^2]=√7/4。
cosA/sinA+cosC/sinC
=(cosAsinC+sinAcosC)/(sinAsinC)
=sin(A+C)/(sinB)^2
=sinB/(sinB)^2
=1/sinB
=4√7/4