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

Another way to look at it is just compute the tangents in terms of the sides of the triangle $\triangle ABC$.
From Law of Cosines we have that $\cos A=\dfrac{b^2+c^2-a^2}{2bc}$, then follows
$$2\sin^2\dfrac{A}{2}=1-\cos A=\dfrac{a^2-(b-c)^2}{2bc}=\dfrac{(a+b-c)(a-b+c)}{2bc}=\dfrac{2(p-c)(p-b)}{bc}$$
where $2p=a+b+c$. Similarly, $$2\cos^2\dfrac{A}{2}=1+\cos A=\dfrac{(b+c)^2-a^2}{2bc}=\dfrac{(-a+b+c)(a+b+c)}{2bc}=\dfrac{2p(p-a)}{bc}.$$
Then, $\tan\dfrac{A}{2}=\sqrt{\dfrac{(p-b)(p-c)}{p(p-a)}}$. Using symmetry we are able to show that $\tan\dfrac{C}{2}=\sqrt{\dfrac{(p-a)(p-b)}{p(p-c)}}$.
Finally, since the sides of $\triangle ABC$ are in $AP$, then $a=b-k$ and
$c=b+k$ for some $k\in\Bbb R$,
$$\tan\dfrac{A}{2}\tan\dfrac{C}{2}=\sqrt{\dfrac{(p-b)(p-c)}{p(p-a)}}\sqrt{\dfrac{(p-a)(p-b)}{p(p-c)}}=\dfrac{p-b}{p}=\dfrac{\dfrac{3b}{2}-b}{\dfrac{3b}{2}}=\dfrac13$$
