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 $\cot \frac{A}{2} \cot \frac{C}{2} = \sqrt{\frac{s(s-a) \cdot s(s-c)}{(s-b)(s-c) \cdot (s-a)(s-b)}} = \sqrt{\frac{s^2}{(s-b)^2}} = \frac{s}{s-b}$.Since $s = \frac{a+b+c}{2}$,we have $2s = a+b+c$.Thus,$\frac{s}{s-b} = \frac{2s}{2s-2b} = \frac{a+b+c}{a+b+c-2b} = \frac{(a+c)+b}{(a+c)-b}$.Given $a+c=5b$,substituting this into the expression gives $\frac{5b+b}{5b-b} = \frac{6b}{4b} = \frac{3}{2}$.

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

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