In a triangle ABC,with usual notations,if 3b | vedclass

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(B) We know that $\cot \frac{A}{2} = \sqrt{\frac{s(s-a)}{(s-b)(s-c)}}$ and $\cot \frac{C}{2} = \sqrt{\frac{s(s-c)}{(s-a)(s-b)}}$.Multiplying these,we

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$2$

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(B) We know that $\cot \frac{A}{2} = \sqrt{\frac{s(s-a)}{(s-b)(s-c)}}$ and $\cot \frac{C}{2} = \sqrt{\frac{s(s-c)}{(s-a)(s-b)}}$.Multiplying these,we get $\cot \frac{A}{2} \cdot \cot \frac{C}{2} = \sqrt{\frac{s(s-a)}{(s-b)(s-c)} \cdot \frac{s(s-c)}{(s-a)(s-b)}} = \frac{s}{s-b}$.Given $3b = a + c$,we know $2s = a + b + c = 3b + b = 4b$,so $s = 2b$.Substituting $s = 2b$ into the expression,we get $\frac{2b}{2b - b} = \frac{2b}{b} = 2$.

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

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