<s>[INST] 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.
 [/INST] 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}

</s>