tangents and triangle

In a triangle, $C = 180^\circ - A - B$ , therefore, $\sin C = \sin \left(180^\circ - A - B \right) = \sin (A+B)$ . Therefore,

$\begin{aligned} \frac{\tan A - \tan B}{\tan A + \tan B} & = \color{#3D99F6}{\frac{c-b}{c}} \quad \quad \small \color{#3D99F6}{\text{By sine rule: } \frac{b}{\sin B} = \frac{c}{\sin C} = \frac{c}{\sin (A+B)}} \\ \frac{\sin A \cos B - \sin B \cos A}{\sin A \cos B + \sin B \cos A} & = \color{#3D99F6}{\frac{\sin (A+B) - \sin B}{\sin (A+B)}} \\ \frac{\sin A \cos B - \sin B \cos A}{\sin (A+B)} & = \frac{\sin (A+B) - \sin B}{\sin (A+B)} \\ \sin A \cos B - \sin B \cos A & = \sin (A+B) - \sin B \\ \sin A \cos B - \sin B \cos A & = \sin A \cos B + \sin B \cos A - \sin B \\ 2 \sin B \cos A & = \sin B \\ \Rightarrow \cos A & = \frac{1}{2} \quad \Rightarrow A = 60^\circ \\ \Rightarrow 2 \tan^2 A & = 2 \tan^2 60^\circ = 2(\sqrt{3})^2 = \boxed{6} \end{aligned}$