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

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(C) 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,w

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$2$

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(C) 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:$cot\, \frac{B}{2} \cdot cot\, \frac{C}{2} = \sqrt{\frac{s(s-b)}{(s-a)(s-c)} \cdot \frac{s(s-c)}{(s-a)(s-b)}} = \sqrt{\frac{s^2}{(s-a)^2}} = \frac{s}{s-a}$.Given $b + c = 3a$,the perimeter $2s = a + b + c = a + 3a = 4a$,so $s = 2a$.Substituting $s = 2a$ into the expression:$\frac{s}{s-a} = \frac{2a}{2a-a} = \frac{2a}{a} = 2$.

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

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