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In $\triangle \mathrm{ABC}$, if a $\cos ^2 \frac{\mathrm{C}}{2}+\cos ^2 \frac{\mathrm{A}}{2}=\frac{3 \mathrm{~b}}{2}$, then $\mathrm{a}+\mathrm{c}: \mathrm{b}=$
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The correct answer is:
$2: 1$
$\begin{aligned} & \text {Given } a \cos ^2 \frac{C}{2}+c \cos ^2 \frac{A}{2}=\frac{3 b}{2} \\ & \Rightarrow \frac{a(1+\cos C)+c(1+\cos A)}{2}=\frac{3 b}{2} \\ & \Rightarrow a+c+a \cos C+c \cos A=3 b \\ & \Rightarrow a+c+b=3 b \Rightarrow \frac{a+c}{b}=\frac{2}{1}\end{aligned}$
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