If sides $a$, $b$, $c$ of $\triangle ABC$ are in arithmetic progression, then $3\tan\frac{A}{2}\tan\frac {C}{2}=1$ 
If sides $a$, $b$, $c$ of $\triangle ABC$ (with $a$ opposite $A$, etc) are in arithmetic progression, then prove that 
  $$3\tan\frac{A}{2}\tan\frac{C}{2}=1$$

My attempt:
$a$, $b$, $c$ are in arithmetic progression, so
$$\begin{align}
2b&=a+c \\[4pt]
2\sin B &= \sin A+ \sin C \\[4pt]
2\sin(A+C) &=2\sin\frac {A+C}{2}\;\cos\frac{A-C}{2} \\[4pt]
2\sin\frac{A+C}{2}\;\cos\frac{A+C}{2}&=\sin\frac{A+C}{2}\;\cos\frac{A-C}{2} \\[4pt]
2\cos\frac{A+C}{2}&=\cos\frac{A-C}{2}
\end{align}$$

Hint:
$$\dfrac21=\dfrac{\cos\dfrac{A-C}2}{\cos\dfrac{A+C}2}$$
Apply Componendo and Dividendo
$$\dfrac{2+1}{2-1}=?$$
