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.

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
