prove $a,b,c$ in A.P if $\tan\dfrac{A}{2}=\dfrac{5}{6}$ and $\tan\dfrac{C}{2}=\dfrac{2}{5}$ 
In $\Delta ABC$, if $\tan\dfrac{A}{2}=\dfrac{5}{6}$ and $\tan\dfrac{C}{2}=\dfrac{2}{5}$, then prove that the sides $a,b,c$ are in A.P.

My Attempt
$$
\sin A=\frac{2.5}{6}.\frac{36}{61}=\frac{60}{61}\\
\sin C=\frac{2.2}{5}.\frac{25}{29}=\frac{20}{29}\\
$$
it is solved in my reference some fomula involving $2s=a+b+c$, can I prove it using the basic known properties of triangles ?

It is $$\tan(\alpha/2)=\frac{r}{s-a}$$ and $$\tan(\gamma/2)=\frac{r}{s-c}$$ where $$s=\frac{a+b+c}{2}$$ so we get $$\frac{5}{6}(s-a)=\frac{2}{5}(s-c)$$ 
Can you finish now?
