2. Angles $\alpha$ and $\beta$ are less than $90^{\circ}$, otherwise if w.l.o.g. $\alpha \geq 90^{\circ}$ we have $\tan (\gamma / 2) \cdot(a \tan \alpha+b \tan \beta)<b \tan (\gamma / 2) \tan \beta \leq b \tan (\gamma / 2) \cot (\gamma / 2)=$ $b<a+b$. Since $a \geq b \Leftrightarrow \tan a \geq \tan b$, Chebyshev's inequality gives $a \tan \alpha+b \tan \beta \geq(a+b)(\tan \alpha+\tan \beta) / 2$. Due to the convexity of the $\tan$ function we also have $(\tan \alpha+\tan \beta) / 2 \geq \tan [(\alpha+\beta) / 2]=\cot (\gamma / 2)$. Hence we have
$$\begin{aligned} \tan \frac{\gamma}{2}(a \tan \alpha+b \tan \beta) & \geq \frac{1}{2} \tan \frac{\gamma}{2}(a+b)(\tan \alpha+\tan \beta) \\ & \geq \tan \frac{\gamma}{2}(a+b) \cot \frac{\gamma}{2}=a+b \end{aligned}$$
The equalities can hold only if $a=b$. Thus the triangle is isosceles.