Question

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

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

Answer 1

The correct option is A

$2$

$\cot \frac{B}{2}=\sqrt{\frac{s(s-b)}{(s-a)(s-c)}}$

$\cot \frac{C}{2}=\sqrt{\frac{s(s-c)}{(s-a)(s-b)}}$

$\therefore \cot \frac{B}{2}\cot \frac{C}{2}=\sqrt{\frac{s(s-b)}{(s-a)(s-c)}\sqrt{\frac{s(s-c)}{(s-a)(s-b)}}}$

$\cot \frac{B}{2}\cot \frac{C}{2}=\sqrt{\frac{s^2(s-b)(s-c)}{(s-a{)}^2(s-c\left)\right(s-b)}}$

$\cot \frac{B}{2}\cot \frac{C}{2}=\sqrt{\frac{s^2}{(s-a{)}^2}}$

$\cot \frac{B}{2}\cot \frac{C}{2}=\frac{s}{(s-a)}$

$\text{Given:}b+c=3a$

$s=\frac{a+c+b}{2}=\frac{a+3a}{2}=2a$

$\therefore \cot \frac{B}{2}\cot \frac{C}{2}=\frac{2a}{2a-a}=\frac{2a}{a}=2$