△ABC中,b2=ac,cosB=3/4 1)cotA+cotC的值 2)设ac=2,求a+c
问题描述：
△ABC中,b2=ac,cosB=3/4 1)cotA+cotC的值 2)设ac=2,求a+c
答
(1)由b^2=ac及正弦定理得(sinB)^2=sinA*sinC
又由cosB=3/4得sinB=√7/4
所以cotA+cotC=(sinCcosA+cosCsinA)/(sinA*sinC)
=sin(A+C)/(sinB)^2
=sin(π-B)/(sinB)^2
=sinB/(sinB)^2
=1/sinB
=4√7/7
(2)由余弦定理得
b^2=a^2+c^2-2accosB
=(a+c)^2-2ac(1+cosB)
=(a+c)^2-2*2*(1+3/4)
=ac=2
∴(a+c)^2=9,a+c=3