Question

If the sides $a,b,c$ of $△ABC$ are in $A.P.$, prove that
$\tan \frac{A}{2}+\tan \frac{C}{2}=\frac{2}{3}\cot \frac{B}{2}.$

Answer 1

By result we have

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

Where $2s=a+b+c=b+2b=3b$

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