sequences and series - If sides $a$, $b$, $c$ of $triangle ABC$ are in arithmetic progression, then $3tanfrac{A}{2}tanfrac {C}{2}=1$

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
$$\begin{align} 2b&=a+c \\[4pt] 2\sin B &= \sin A+ \sin C \\[4pt] 2\sin(A+C) &=2\sin\frac {A+C}{2}\;\cos\frac{A-C}{2} \\[4pt] 2\sin\frac{A+C}{2}\;\cos\frac{A+C}{2}&=\sin\frac{A+C}{2}\;\cos\frac{A-C}{2} \\[4pt] 2\cos\frac{A+C}{2}&=\cos\frac{A-C}{2} \end{align}$$

Answer

Expand your last line: $$2\left(\cos\frac A2\cos\frac C2 - \sin\frac A2\sin\frac C2\right)=\left(\cos\frac A2\cos\frac C2 +\sin\frac A2\sin\frac C2\right)$$
and your result is immediate after a cancellation.