In a triangle ABC,if 3a = b + c,then cot frac | vedclass

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(B) Given,$3a = b + c$ ... $(i)$Let $s$ be the semi-perimeter of $	riangle ABC$,so $s = \frac{a + b + c}{2}$.Substituting the value from $(i)$,we get

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$2$

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(B) Given,$3a = b + c$ ... $(i)$Let $s$ be the semi-perimeter of $	riangle ABC$,so $s = \frac{a + b + c}{2}$.Substituting the value from $(i)$,we get $s = \frac{a + 3a}{2} = \frac{4a}{2} = 2a$.We know that $\cot \frac{B}{2} = \sqrt{\frac{s(s-b)}{(s-a)(s-c)}}$ and $\cot \frac{C}{2} = \sqrt{\frac{s(s-c)}{(s-a)(s-b)}}$.Therefore,$\cot \frac{B}{2} \cot \frac{C}{2} = \sqrt{\frac{s(s-b)}{(s-a)(s-c)} \cdot \frac{s(s-c)}{(s-a)(s-b)}} = \sqrt{\frac{s^2}{(s-a)^2}} = \frac{s}{s-a}$.Substituting $s = 2a$,we get $\frac{2a}{2a - a} = \frac{2a}{a} = 2$.

In a $\triangle ABC$,if $3a = b + c$,then $\cot \frac{B}{2} \cot \frac{C}{2} =$

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