In a triangle ABC,if a+c=5b,then cot frac{A}{ | vedclass

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(C) 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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$\frac{3}{2}$

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(C) 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 $\cot \frac{A}{2} \cot \frac{C}{2} = \sqrt{\frac{s(s-a)}{(s-b)(s-c)} 	imes \frac{s(s-c)}{(s-a)(s-b)}} = \sqrt{\frac{s^2}{(s-b)^2}} = \frac{s}{s-b}$.Given $a+c=5b$,the semi-perimeter $s = \frac{a+b+c}{2} = \frac{5b+b}{2} = 3b$.Substituting $s = 3b$ into the expression,we get $\frac{s}{s-b} = \frac{3b}{3b-b} = \frac{3b}{2b} = \frac{3}{2}$.

In a $\triangle ABC$,if $a+c=5b$,then $\cot \frac{A}{2} \cot \frac{C}{2} =$

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