trigonometry - Trigonometric Arithmetic Progression

If $a$, $b$, $c$ are in arithmetic progression, prove that
$$\cos A \cot\frac{A}{2} \qquad \cos B \cot \frac{B}{2} \qquad \cos C \cot\frac{C}{2}$$
are in arithmetic progression, too.

Here, $a$, $b$, $c$ represent the sides of a triangle and $A$, $B$, $C$ are the opposite angles of the triangle.

Answer

For better clarity, I'm adding another proof that $\displaystyle\cot\frac A2,\cot\frac B2,\cot\frac C2$ are also in AP if $a,b,c$ are so.

We have $\displaystyle00$

So, $\displaystyle\cot\frac C2=\frac1{\tan\frac C2}=+\sqrt{\frac{1+\cos A}{1-\cos A}}$

Using Law of Cosines and on simplification, $\displaystyle\cot\frac C2=+\sqrt{\frac{s(s-c)}{(s-b)(s-a)}}$ where $2s=a+b+c$

$\displaystyle\cot\frac A2,\cot\frac B2,\cot\frac C2$ will be in AP

$\displaystyle\iff\sqrt{\frac{s(s-c)}{(s-b)(s-a)}}+\sqrt{\frac{s(s-a)}{(s-b)(s-c)}}=\sqrt{\frac{s(s-b)}{(s-c)(s-a)}}$

$\displaystyle\iff s-a+s-c=2(s-b)\iff a+c=2b$