Question

If $B+C=3A$, prove that $\cot \frac{1}{2}B\cot \frac{1}{2}C=2$.

Answer 1

Since $b+c=3a$ we have $\sin B+\sin C=3\sin A$
or $2\sin \frac{B+C}{2}\cos \frac{B-C}{2}=6\sin \frac{A}{2}\cos \frac{A}{2}$
or $\cos \frac{B-C}{2}=3\sin \frac{A}{2}[\because \sin \frac{B+C}{2}=\cos \frac{A}{2}]$
or $\cos \frac{B-C}{2}=3\cos \frac{B+C}{2}$
or $\left\{\cos \frac{B-C}{2}\right\}/\left\{\cos \frac{B+C}{2}\right\}=\frac{3}{1}$
Apply $C$ and $D\therefore \cot \left(B/2\right)\cot \left(C/2\right)=2$
Converse Method:
On putting the value of $\left(B/2\right)$ and $\left(C/2\right)$ in $R.H.S$ you will get
$s/(s-a)=2$ or $s=2a$
or $a+b+c=4a\therefore b+c=3a$.
This proves the converse.