Question

If the sides $a,b,c$ of a triangle are in arithmetic progression, then the value of $\tan \frac{A}{2}+\tan \frac{C}{2}$ in terms of $\cot \frac{B}{2}$ is $\frac{a}{b}\cot \frac{B}{2}$ then $a+b=$

Answer 1

Since th sides $a,b,c$ are in A.P.

$\therefore 2b=a+c$ ...(1)

Now, $\tan \frac{A}{2}+\tan \frac{C}{2}=\frac{\sin \frac{A}{2}}{\cos \frac{A}{2}}+\frac{\sin \frac{C}{2}}{\cos \frac{C}{2}}$

$=\frac{\sin \frac{A}{2}\cos \frac{C}{2}+\cos \frac{A}{2}\sin \frac{C}{2}}{\cos \frac{A}{2}\cos \frac{C}{2}}=\frac{\sin \left(\frac{A+C}{2}\right)}{\cos \frac{A}{2}\cos \frac{C}{2}}$ ...(2)

$(\because A+B+C={180}^0,\frac{A+C}{2}={90}^0-\frac{B}{2})$

Now, $\cos \frac{A}{2}\cos \frac{C}{2}={\left(\frac{s(s-a)s(s-c)}{bc.ba}\right)}^{\frac{1}{2}}=\frac{s}{b}{\left(\frac{(s-a)(s-c)}{ac}\right)}^{\frac{1}{2}}$

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

Hence from (2), we have

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