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.
 A: 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 $\displaystyle0<C<\pi\iff 0<\frac C2<\frac\pi2\implies\cot\frac C2>0$
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 
$\displaystyle\iff\sqrt{\frac{s(s-c)}{(s-b)(s-a)}}+\sqrt{\frac{s(s-a)}{(s-b)(s-c)}}=\sqrt{\frac{s(s-b)}{(s-c)(s-a)}}$
$\displaystyle\iff s-a+s-c=2(s-b)\iff a+c=2b$
A: $$\cos B\cot\frac B2=\left(1-2\sin^2\frac B2\right)\cot\frac B2=\cot\frac B2-\sin A$$
Clearly using Law of Sines,  $\sin A,\sin B,\sin C$ are in AP
So,  it is sufficient to show that $\displaystyle\cot\frac A2,\cot\frac B2,\cot\frac C2$ are also in AP
We have $$a+c=2b\iff \sin A+\sin C=2\sin B$$
Now using Prosthaphaeresis Formula, $$2\sin\frac{A+C}2\cos\frac{A-C}2=2(2\sin\frac B2\cos\frac B2)$$
$$\iff\cos\frac{A-C}2=2\sin\frac B2\ \ \ \ (1)$$ 
Again, $\displaystyle\cot\frac A2+\cot\frac C2=2\cot\frac B2$
$$\iff\frac{\sin\frac{A+C}2}{\cos\frac A2\cos\frac C2}=2\frac{\cos\frac B2}{\sin\frac B2}$$
$$\iff\sin\frac B2=2\cos\frac A2\cos\frac C2=\cos\frac{A-C}2-\cos\frac{A+C}2$$
$$\iff\sin\frac B2=\cos\frac{A-C}2-\sin\frac B2$$ which is same as $(1)$
