In a Delta ABC, if 3a = b + c, then the value | vedclass

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(B) We know that $\cot \frac{B}{2} = \sqrt{\frac{s(s-b)}{(s-a)(s-c)}}$ and $\cot \frac{C}{2} = \sqrt{\frac{s(s-c)}{(s-a)(s-b)}}.$ Multiplying these,we

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$2$

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(B) We know that $\cot \frac{B}{2} = \sqrt{\frac{s(s-b)}{(s-a)(s-c)}}$ and $\cot \frac{C}{2} = \sqrt{\frac{s(s-c)}{(s-a)(s-b)}}.$ Multiplying these,we get $\cot \frac{B}{2} \cot \frac{C}{2} = \sqrt{\frac{s(s-b)}{(s-a)(s-c)} 	imes \frac{s(s-c)}{(s-a)(s-b)}} = \sqrt{\frac{s^2}{(s-a)^2}} = \frac{s}{s-a}.$ Given $3a = b + c,$ we add $a$ to both sides to get $a + b + c = 4a,$ which implies $2s = 4a,$ so $s = 2a.$ Substituting $s = 2a$ into the expression,we get $\frac{2a}{2a - a} = \frac{2a}{a} = 2.$

In a $\Delta ABC,$ if $3a = b + c,$ then the value of $\cot \frac{B}{2} \cot \frac{C}{2}$ is

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