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In a $\triangle A B C$, if $a+c=5 b$, then $\cot \frac{A}{2} \cot \frac{C}{2}$ is equal to
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$\frac{3}{2}$
$\begin{aligned} & \text { LHS } \Rightarrow \cot \frac{A}{2} \cdot \cot \frac{C}{2}=\sqrt{\frac{s(s-a)}{(s-b)(s-c)}} \sqrt{\frac{s(s-c)}{(s-a)(s-b)}} \\ & =\sqrt{\frac{s(s-a)(s-c)}{(s-b)(s-c)(s-a)(s-b)}}=\frac{s}{s-b}=\frac{2 s}{2 s-2 b} \\ & =\frac{a+b+c}{a+b+c-2 b}=\frac{5 b+b}{5 b-b}=\frac{6 b}{4 b}=\frac{3}{2}\end{aligned}$
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