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

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(C) Using the formula for $\cot \frac{B}{2} \cdot \cot \frac{C}{2}$,we have: $\cot \frac{B}{2} \cdot \cot \frac{C}{2} = \sqrt{\frac{s(s-b)}{ac}} \cdot

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$2$

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(C) Using the formula for $\cot \frac{B}{2} \cdot \cot \frac{C}{2}$,we have: $\cot \frac{B}{2} \cdot \cot \frac{C}{2} = \sqrt{\frac{s(s-b)}{ac}} \cdot \sqrt{\frac{s(s-c)}{ab}} \cdot \frac{1}{\sin(B/2)\sin(C/2)} \dots$ $A$ simpler approach is using the identity $\cot \frac{B}{2} \cot \frac{C}{2} = \frac{s}{s-a}$. Given $3a = b + c$,we know $2s = a + b + c = a + 3a = 4a$,so $s = 2a$. Substituting this into the identity: $\cot \frac{B}{2} \cot \frac{C}{2} = \frac{2a}{2a - a} = \frac{2a}{a} = 2$. Thus,the correct option is $C$.

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

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