<s>[INST] 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 ?
 [/INST] You can first deduce
$$
\tan\frac{B}{2}=\tan\left(\frac{\pi}{2}-\frac{A+C}{2}\right)=\cot\frac{A+C}{2}=
\frac{1-\tan\frac{A}{2}\tan\frac{C}{2}}{\tan\frac{A}{2}+\tan\frac{C}{2}}=\frac{2/3}{37/30}
=\frac{20}{37}
$$
Therefore
$$
\sin A=\frac{2(5/6)}{1+25/36}=\frac{60}{61}
$$
Similarly,
$$
\sin B=\frac{1480}{1769}\qquad \sin C=\frac{20}{29}
$$
By the sine law,
$$
\frac{a+c}{2}=\frac{b}{2\sin B}(\sin A+\sin C)=
b\frac{1769}{2960}\left(\frac{60}{61}+\frac{20}{29}\right)=b
$$
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