In a $ riangle A B C$, if $3 a=b+c$, then $\cot \frac{B}{2} \cot \frac{C}{2}=$

Question

In a $	riangle A B C$, if $3 a=b+c$, then $\cot \frac{B}{2} \cot \frac{C}{2}=$

Aicharya · Accepted Answer

Let $a, b, c$ are the sides of the $	riangle A B C$ and $s$ is the semi-perimeter $ s=\frac{a+b+c}{2} $ Put the value of Eq. (i), we get $ s=\frac{a+3 a}{2}=\frac{4 a}{2}=2 a $ Now, $\cot \frac{B}{2} \cot \frac{C}{2}$ $ \begin{gathered} =\sqrt{\frac{s(s-b)}{(s-a)(s-c)}} \cdot \sqrt{\frac{s(s-c)}{(s-a)(s-b)}} \ \left\{\because \cot \frac{A}{2}=\sqrt{\frac{s(s-a)}{(s-b)(s-c)}}ight\} \end{gathered} $ Put the value of $s$, we get $ \begin{aligned} & =\sqrt{\frac{2 a(2 a-b)}{(2 a-a)(2 a-c)}} \cdot \sqrt{\frac{2 a(2 a-c)}{(2 a-a)(2 a-b)}} \ & =\sqrt{\frac{2 a(2 a-b)}{a(2 a-c)} 	imes \frac{2 a(2 a-c)}{a(2 a-b)}}=\sqrt{4}=2 \end{aligned} $

In a $\triangle A B C$, if $3 a=b+c$, then $\cot \frac{B}{2} \cot \frac{C}{2}=$

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