Question

If the sides $a,b,c$ of a triangle are in A.P., then find the value of $\tan \frac{A}{2}+\tan \frac{C}{2}$ in terms of $\cot \left(\frac{B}{2}\right)$

Answer 1

As $a,b,c$ are in AP,

$a+c=2b$

Consider, $\tan \frac{A}{2}+\tan \frac{C}{2}$

We have the formula $\tan \frac{A}{2}=\sqrt{\frac{(s-b)(s-c)}{s(s-a)}}$

Substituting, we get,

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

$=\sqrt{\frac{(s-b)}{s}}[\sqrt{\frac{(s-c)}{(s-a)}}+\sqrt{\frac{(s-a)}{(s-c)}}]$

$=\sqrt{\frac{(s-b)}{s}}\left[\frac{(s-c)+(s-a)}{\sqrt{(s-a)(s-c)}}\right]$

$=\sqrt{\frac{(s-b)}{s}}\left[\frac{(\frac{a+b+c}{2}-c)+(\frac{a+b+c}{2}-a)}{\sqrt{(s-a)(s-c)}}\right]$ --------[$s=\frac{a+b+c}{2}$]

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

$=\frac{b}{s}\cot \frac{B}{2}$

We have, $s=\frac{a+b+c}{2}$ and $a+c=2b$

$\therefore s=\frac{2b+b}{2}=\frac{3b}{2}$

Substitute for $s$ in

$\tan \frac{A}{2}+\tan \frac{C}{2}=\frac{b}{s}\cot \frac{B}{2}$

We get,

$\tan \frac{A}{2}+\tan \frac{C}{2}=\frac{2}{3}\cot \frac{B}{2}$