Question

If $ABC$ is triangle and $\tan \frac{A}{2},\tan \frac{B}{2},\tan \frac{C}{2}$ are in H.P the find the minimum value of $\cot \frac{B}{2}$

Answer 1

$ABC$ is a triangle, therefore $A+B+C=\pi$
$\Rightarrow \cot \frac{A}{2}\cos \frac{B}{2}\cot \frac{C}{2}=\cot \frac{A}{2}+\cot \frac{B}{2}+\cot \frac{C}{2}$
But $\tan \frac{A}{2},\tan \frac{B}{2},\tan \frac{C}{2}\to H.P$
$\Rightarrow \cot \frac{A}{2}+\cot \frac{B}{2}+\cot \frac{C}{2}\to A.P$
$\Rightarrow \cot \frac{A}{2}+\cot \frac{C}{2}=2\cot \frac{B}{2}$
Therefore
$\cot \frac{A}{2}.\cot \frac{B}{2}.\cot \frac{C}{2}=3\cot \frac{B}{2}$
$\therefore \cot \frac{A}{2}.\cot \frac{C}{2}=3$
Now $\frac{\cos \frac{A}{2}+\cot \frac{C}{2}}{2}\ge \sqrt{\cos \frac{A}{2}\cot \frac{C}{2}}\Rightarrow \frac{2\cot \frac{B}{2}}{2}\ge \sqrt{3}$
$\therefore \cot \frac{B}{2}\ge \sqrt{3}$