In any $\triangle ABC$, if $\;$ $\tan \left(\dfrac{A}{2}\right) = \dfrac{5}{6}$, $\;$ $\tan \left(\dfrac{B}{2}\right) = \dfrac{20}{37}$, $\;$ find $\;$ $\tan \left(\dfrac{C}{2}\right)$ and prove that in this triangle $\;$ $a + c = 2b$.
Given: $\;$ $\tan \left(\dfrac{A}{2}\right) = \dfrac{5}{6}$, $\;$ $\tan \left(\dfrac{B}{2}\right) = \dfrac{20}{37}$
In $\triangle ABC$, $\;$ $A + B + C = \pi$
i.e. $\;$ $C = \pi - \left(A + B\right)$
i.e. $\;$ $\dfrac{C}{2} = \dfrac{\pi}{2} - \left(\dfrac{A}{2} + \dfrac{B}{2}\right)$
$\therefore \;$ $\tan \left(\dfrac{C}{2}\right) = \tan \left[\dfrac{\pi}{2} - \left(\dfrac{A}{2} + \dfrac{B}{2}\right)\right] = \cot \left(\dfrac{A}{2} + \dfrac{B}{2}\right)$
Now, $\cot \left(\alpha + \beta\right) = \dfrac{1 - \tan \alpha \tan \beta}{\tan \alpha + \tan \beta}$
$\therefore \;$ We have
$\begin{aligned}
\tan \left(\dfrac{C}{2}\right) & = \cot \left(\dfrac{A}{2} + \dfrac{B}{2}\right) \\\\
& = \dfrac{1 - \tan \left(\dfrac{A}{2}\right) \times \tan \left(\dfrac{B}{2}\right)}{\tan \left(\dfrac{A}{2}\right) + \tan \left(\dfrac{B}{2}\right)} \\\\
& = \dfrac{1 - \dfrac{5}{6} \times \dfrac{20}{37}}{\dfrac{5}{6} + \dfrac{20}{37}} \\\\
& = \dfrac{122}{305}
\end{aligned}$
i.e. $\;$ $\tan \left(\dfrac{C}{2}\right) = \dfrac{2}{5}$
For this given triangle,
To Prove That (TPT) $\;$ $a + c = 2b$
i.e. $\;$ $TPT \;\;$ $\dfrac{a}{2 R} + \dfrac{c}{2 R} = 2 \times \dfrac{b}{2 R}$ $\;\;\;$ [$R$ is the circumradius of $\triangle ABC$]
$\left[\text{Note: Sine rule: } \dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C} = 2 R \right.$
$\left. \implies \dfrac{a}{2R} = \sin A, \; \dfrac{b}{2 R} = \sin B, \; \dfrac{c}{2 R} = \sin C \right]$
i.e. $\;$ $TPT \;\;$ $\sin A + \sin C = 2 \sin B$
$\left[\text{Note: } \sin \theta = \dfrac{2 \tan \left(\dfrac{\theta}{2}\right)}{1 + \tan^2 \left(\dfrac{\theta}{2}\right)}\right]$
i.e. $\;$ $TPT \;\;$ $\dfrac{2 \tan \left(\dfrac{A}{2}\right)}{1 + \tan^2 \left(\dfrac{A}{2}\right)} + \dfrac{2 \tan \left(\dfrac{C}{2}\right)}{1 + \tan^2 \left(\dfrac{C}{2}\right)} = 2 \times \dfrac{2 \tan \left(\dfrac{B}{2}\right)}{1 + \tan^2 \left(\dfrac{B}{2}\right)}$
i.e. $\;$ $TPT \;\;$ $\dfrac{\tan \left(\dfrac{A}{2}\right)}{1 + \tan^2 \left(\dfrac{A}{2}\right)} + \dfrac{\tan \left(\dfrac{C}{2}\right)}{1 + \tan^2 \left(\dfrac{C}{2}\right)} = \dfrac{2 \tan \left(\dfrac{B}{2}\right)}{1 + \tan^2 \left(\dfrac{B}{2}\right)}$
i.e. $\;$ $TPT \;\;$ $\dfrac{\dfrac{5}{6}}{1 + \dfrac{25}{36}} + \dfrac{\dfrac{2}{5}}{1 + \dfrac{4}{25}} = \dfrac{2 \times \dfrac{20}{37}}{1 + \dfrac{400}{1369}}$
i.e. $\;$ $TPT \;$ $\dfrac{30}{61} + \dfrac{10}{29} = \dfrac{1480}{1769}$
i.e. $\;$ $TPT \;$ $\dfrac{1480}{1769} = \dfrac{1480}{1769}$ $\;$ which is true.
$\therefore \;$ $a + c = 2b$
Hence proved.