(D) 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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(D) 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 semi-perimeter $s = \frac{a+b+c}{2} = \frac{a+3a}{2} = 2a$.Substituting $s = 2a$ into the expression,we get $\frac{s}{s-a} = \frac{2a}{2a-a} = \frac{2a}{a} = 2$.

Class 11 Mathematics Trigonometrical Equations Relation between sides and angles, Solutions of triangles

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

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A
$3$
B
$1$
C
$4$
D
$2$

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

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