<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] Writing $A_2$ for $A/2$, etc, and noting that $A_2+B_2+C_2 = \pi/2$, we have
$$\begin{align}
\tan A_2-\tan B_2 &= \tan B_2 - \tan C_2 \\[6pt]
\frac{\sin A_2 \cos B_2 - \cos A_2 \sin B_2}{\cos A_2 \cos B_2}
&=
\frac{\sin B_2 \cos C_2 - \cos B_2 \sin C_2}{\cos B_2 \cos C_2} \\[6pt]
\sin(A_2-B_2)\cos C_2 &= \sin(B_2-C_2)\cos A_2 \\[6pt]
\sin(A_2-B_2)\sin(A_2+B_2) &= \sin(B_2-C_2)\sin(B_2+C_2) \\[6pt]
\frac12\left(\cos 2B_2 - \cos 2 A_2\right) &= \frac12\left(\cos 2C_2-\cos 2B_2\right) \\[6pt]
\cos B - \cos A &= \cos C - \cos B
\end{align}$$
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