Question

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

• A. $\sqrt{3}$
• B. $1$
• C. $2$
• D. $3$

Answer 1

Answer: (C)

$\because$ $3a=b+c$
$\Rightarrow$ $\frac{b+c}{a}=3$
Applying sine rule, we get
$\frac{\sin B+\sin C}{\sin A}=3$
$\Rightarrow$ $\frac{2\sin \frac{B+C}{2}\cos \frac{B-C}{2}}{2\sin \frac{A}{2}\cos \frac{A}{2}}=3$
$\Rightarrow$ $\frac{\cos \frac{A}{2}\cos \frac{B-C}{2}}{\cos \left(\frac{B+C}{2}\right)\cos \frac{A}{2}}=3$
$\Rightarrow$ $\cos \frac{B}{2}\cos \frac{C}{2}+\sin \frac{B}{2}\sin \frac{C}{2}=$
$3\left[\cos \frac{B}{2}\cos \frac{C}{2}-\sin \frac{B}{2}\sin \frac{C}{2}\right]$
$\Rightarrow$ $2\cos \frac{B}{2}\cos \frac{C}{2}=4\sin \frac{B}{2}\sin \frac{C}{2}$
$\Rightarrow$ $\cot \frac{B}{2}\cot \frac{C}{2}=2$
Alternate Solution:
We have, $3a=b+c$ ...(i)
Now, $\cot \frac{B}{2}\cot \frac{C}{2}$
$\Rightarrow$ $\sqrt{\frac{s(s-b)}{(s-a)(s-c)}}.\sqrt{\frac{s(s-c)}{(s-a)(s-b)}}$
$=\frac{s}{s-a}=\frac{\frac{a+b+c}{2}}{\frac{a+b+c}{2}-a}$
$=\frac{a+b+c}{-a+b+c}$
$=\frac{a+3a}{-a+3a}$ [from (i)]
$=\frac{4a}{2a}=2$