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

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(B) We know the half-angle formulas for a triangle: $\cot \frac{B}{2} = \sqrt{\frac{s(s-b)}{(s-a)(s-c)}}$ and $\cot \frac{C}{2} = \sqrt{\frac{s(s-c)}{

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$2$

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(B) We know the half-angle formulas for a triangle: $\cot \frac{B}{2} = \sqrt{\frac{s(s-b)}{(s-a)(s-c)}}$ and $\cot \frac{C}{2} = \sqrt{\frac{s(s-c)}{(s-a)(s-b)}}$.Multiplying these,we get $\cot \frac{B}{2} \cot \frac{C}{2} = \sqrt{\frac{s(s-b)}{(s-a)(s-c)} \cdot \frac{s(s-c)}{(s-a)(s-b)}} = \sqrt{\frac{s^2}{(s-a)^2}} = \frac{s}{s-a}$.Since $s = \frac{a+b+c}{2}$,we have $2s = a+b+c$.Thus,$\frac{s}{s-a} = \frac{2s}{2s-2a} = \frac{a+b+c}{a+b+c-2a} = \frac{a+b+c}{b+c-a}$.Given $b+c = 3a$,we substitute this into the expression:$\frac{a + 3a}{3a - a} = \frac{4a}{2a} = 2$.

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

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