In a triangle, if $\tan(A/2)$, $\tan(B/2)$, $\tan(C/2)$ are in arithmetic progression, then so are $\cos A$, $\cos B$, $\cos C$ 
In a triangle, if $\tan\frac{A}{2}$, $\tan\frac{B}{2}$, $\tan\frac{C}{2}$ are in arithmetic progression, then show that $\cos A$, $\cos B$, $\cos C$ are in arithmetic progression.

$$2\tan\left(\dfrac{B}{2}\right)=\tan\left(\dfrac{A}{2}\right)+\tan\left(\dfrac{C}{2}\right)$$
$$2\sqrt{\dfrac{(s-a)(s-c)}{s(s-b)}}=\sqrt{\dfrac{(s-b)(s-c)}{s(s-a)}}+\sqrt{\dfrac{(s-a)(s-b)}{s(s-c)}}$$
$$2\sqrt{\dfrac{(s-a)(s-c)(s-b)}{s(s-b)^2}}=\sqrt{\dfrac{(s-a)(s-b)(s-c)}{s(s-a)^2}}+\sqrt{\dfrac{(s-a)(s-b)(s-c)}{s(s-c)^2}}$$
$$\dfrac{2}{s-b}=\dfrac{1}{s-a}+\dfrac{1}{s-c}$$
$$\dfrac{2}{s-b}=\dfrac{s-c+s-a}{(s-a)(s-c)}$$
$$\dfrac{2}{s-b}=\dfrac{b}{(s-a)(s-c)}$$
$$2\left(\dfrac{a+b+c}{2}-a\right)\left(\dfrac{a+b+c}{2}-c\right)=b\left(\dfrac{a+b+c}{2}-b\right)$$
$$2\left(\dfrac{b+c-a}{2}\right)\left(\dfrac{a+b-c}{2}\right)=b\left(\dfrac{a+c-b}{2}\right)$$
$$2\left(\dfrac{b+c-a}{2}\right)\left(\dfrac{a+b-c}{2}\right)=b\left(\dfrac{a+c-b}{2}\right)$$
$$b^2-a^2-c^2+2ac=ba+bc-b^2$$
$$2b^2-a^2-c^2+2ac-ba-bc=0\tag{1}$$
$$\cos B=\dfrac{a^2+c^2-b^2}{2ac}$$
$$\cos A=\dfrac{b^2+c^2-a^2}{2bc}$$
$$\cos C=\dfrac{a^2+b^2-c^2}{2ab}$$
$$\cos A+\cos C=\dfrac{ab^2+ac^2-a^3+a^2c+b^2c-c^3}{2abc}$$
$$\cos A+\cos C=\dfrac{ab+bc+\dfrac{ac^2-a^3+a^2c-c^3}{b}}{2ac}$$
$$\cos A+\cos C=\dfrac{ab+bc+\dfrac{ac(a+c)-(a+c)(a^2+c^2-ac)}{b}}{2ac}$$
Using equation $(1)$, $2ac-a^2-c^2=ba+bc-2b^2$
$$\cos A+\cos C=\dfrac{ab+bc+\dfrac{(a+c)(ba+bc-2b^2)}{b}}{2ac}$$
$$\cos A+\cos C=\dfrac{ab+bc+(a+c)(a+c-2b)}{2ac}$$
$$\cos A+\cos C=\dfrac{ab+bc+a^2+c^2+2ac-2ba-2bc}{2ac}$$
$$\cos A+\cos C=\dfrac{a^2+c^2+2ac-ab-bc}{2ac}$$
Using equation $(1)$, $2ac-ba-bc=a^2+c^2-2b^2$
$$\cos A+\cos C=\dfrac{a^2+c^2+a^2+c^2-2b^2}{2ac}$$
$$\cos A+\cos C=\dfrac{2a^2+2c^2-2b^2}{2ac}$$
$$\cos A+\cos C=2\cdot\dfrac{a^2+c^2-b^2}{2ac}$$
$$\cos A+\cos C=2\cos B$$
Is there any nice way to solve this question, mine goes very long. I tried various methods but this was the only way I was able to prove the required result.
 A: Rewrite the equation as 
$$2\frac{\sin\frac B2}{\cos\frac B2}
=\frac{\sin\frac A2}{\cos\frac A2}+\frac{\sin\frac C2}{\cos\frac C2}
=\frac{\sin\frac {A+C}2}{\cos\frac A2\cos\frac C2}$$
Then, with $A+C = \pi - B$,
$$2\sin\frac B2 \cos\frac A2 \cos\frac C2 = \cos\frac B2\sin\frac {\pi-B}2=\cos^2\frac B2$$
$$2\sin\frac B2(\cos\frac {A+C}2 + \cos\frac {A-C}2) = 1 + \cos B$$
$$2\sin^2\frac B2+ 2\sin\frac B2\cos\frac {A-C}2 = 1 + \cos B$$
$$2\cos\frac {A+C}2\cos\frac {A-C}2 = 2\cos B$$
$$\cos A + \cos C = 2\cos B$$
A: Writing $A_2$ for $A/2$, etc, and noting that $A_2+B_2+C_2 = \pi/2$, we have
$$\begin{align}
\tan A_2-\tan B_2 &= \tan B_2 - \tan C_2 \\[6pt]
\frac{\sin A_2 \cos B_2 - \cos A_2 \sin B_2}{\cos A_2 \cos B_2}
&=
\frac{\sin B_2 \cos C_2 - \cos B_2 \sin C_2}{\cos B_2 \cos C_2} \\[6pt]
\sin(A_2-B_2)\cos C_2 &= \sin(B_2-C_2)\cos A_2 \\[6pt]
\sin(A_2-B_2)\sin(A_2+B_2) &= \sin(B_2-C_2)\sin(B_2+C_2) \\[6pt]
\frac12\left(\cos 2B_2 - \cos 2 A_2\right) &= \frac12\left(\cos 2C_2-\cos 2B_2\right) \\[6pt]
\cos B - \cos A &= \cos C - \cos B
\end{align}$$
A: For $A\ne B,A+B+C=\pi$
using http://mathworld.wolfram.com/ProsthaphaeresisFormulas.html
$$f(B,A)=\dfrac{\tan\dfrac B2-\tan\dfrac A2}{\cos B-\cos A}=-\dfrac1{2\cos\dfrac A2\cos\dfrac B2\cos\dfrac C2}$$
By symmetry, $$f(B,A)=f(C,B)$$
Can you take it from here?
Similarly we can establish $$\dfrac{\cot\dfrac B2-\cot\dfrac A2}{\sin B-\sin C}=-\dfrac1{2\sin\dfrac A2\sin\dfrac B2\sin\dfrac C2}$$ so that a similar problem can be proved
A: Given
\begin{align} 
\tan\tfrac12A&=u-d
,\quad
\tan\tfrac12B=u
,\quad
\tan\tfrac12C=u+d
,\quad
u,d\in\mathbb{R}
\tag{1}\label{1}
.
\end{align} 
We can express $d$ in terms of $u$ using 
known identity for triangles
\begin{align}
\tan\tfrac A2\tan\tfrac B2+
\tan\tfrac B2\tan\tfrac C2+
\tan\tfrac C2\tan\tfrac A2&=1
,\\
(u-d)u+u(u+d)+(u+d)(u-d)=3u^2-d^2 &= 1
,\\
d&=\sqrt{3u^2-1}
\tag{2}\label{2}
,
\end{align}
and \eqref{1} becomes
\begin{align}
\tan\tfrac12A&=u-\sqrt{3u^2-1}
,\quad
\tan\tfrac12B=u
,\quad
\tan\tfrac12C=u+\sqrt{3u^2-1}
\tag{3}\label{3}
.
\end{align}
Since all the tangents must be positive,
we have a condition 
$u\in(\tfrac{\sqrt3}3,\,\tfrac{\sqrt2}2)$,
the endpoints correspond to degenerate solutions:
one corresponds to the equilateral triangle, $d=0$,
and the other corresponds
to the degenerate triangle with $\tan\tfrac12A=0$.
We also know that
\begin{align}
\cos x&=\frac{1-\tan^2\tfrac x2}{1+\tan^2\tfrac x2}
,
\end{align}
so the corresponding three cosines
\begin{align}
\cos A&=
\frac{u-u^3+\sqrt{3u^2-1}}{u(1+u^2)}
,\\
\cos B&=
\frac{1-u^2}{1+u^2}
,\\
\cos C&=
\frac{u-u^3-\sqrt{3u^2-1}}{u(1+u^2)}
\end{align} 
are indeed in arithmetic progression,
\begin{align}
u'-d',\quad &u',\quad u+d'
,\\
u'&=\frac{1-u^2}{1+u^2}
,\qquad
d'=-\frac{\sqrt{3u^2-1}}{u(u^2+1)}
.
\end{align}

Bonus:
a suitable example 
of the triangle $ABC$ with given properties
has integer side lengths 
$a=27$,
$b=32$ and
$c=35$ units, 
and 
\begin{align}
u&=\tfrac{4\sqrt{47}}{47}
,\quad 
d=\tfrac{\sqrt{47}}{47}
,\\
u'&=\tfrac{31}{63}
,\quad
d'=-\tfrac{47}{252}
,\\
\alpha&=
2\arctan(\tfrac{3\sqrt{47}}{47})=\arccos(\tfrac{19}{28})
\approx 47.2679^\circ
,\\
\beta&=
2\arctan(\tfrac{4\sqrt{47}}{47})=\arccos(\tfrac{31}{63})
\approx 60.5237^\circ
,\\
\gamma&=
2\arctan(\tfrac{5\sqrt{47}}{47})=\arccos(\tfrac{11}{36})
\approx 72.2084^\circ
.
\end{align}

