In a triangle ABC , with usual notations, $3 \mathrm{~b}=\mathrm{a}+\mathrm{c}$, then $\cot \frac{\mathrm{A}}{2} \cdot \cot \frac{\mathrm{c}}{2}=$

Using the semi-perimeter $s = \frac{a+b+c}{2}$, the half-angle cotangent formulas yield:

$\cot \frac{A}{2} \cdot \cot \frac{C}{2} = \sqrt{\frac{s(s-a)}{(s-b)(s-c)}} \cdot \sqrt{\frac{s(s-c)}{(s-a)(s-b)}} = \frac{s}{s-b}$

Given $3b = a + c$, the semi-perimeter becomes $s = \frac{(a+c)+b}{2} = \frac{3b+b}{2} = 2b$.

Substituting into the product gives $\frac{2b}{2b-b} = 2$.

Final Answer: B