If sides $a$, $b$, $c$ of $\triangle ABC$ are in arithmetic progression, then $3\tan\frac{A}{2}\tan\frac {C}{2}=1$

Question

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 1

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.

Answer 2

Hint:

$$\dfrac21=\dfrac{\cos\dfrac{A-C}2}{\cos\dfrac{A+C}2}$$

Apply Componendo and Dividendo

$$\dfrac{2+1}{2-1}=?$$

Answer 3

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 $k\in\Bbb R$, $$\tan\dfrac{A}{2}\tan\dfrac{C}{2}=\sqrt{\dfrac{(p-b)(p-c)}{p(p-a)}}\sqrt{\dfrac{(p-a)(p-b)}{p(p-c)}}=\dfrac{p-b}{p}=\dfrac{\dfrac{3b}{2}-b}{\dfrac{3b}{2}}=\dfrac13$$