In $\triangle ABC$, $\;$ prove that $\;$ $\left(a^2 - b^2 + c^2\right) \tan B = \left(a^2 + b^2 - c^2\right) \tan C$
By sine rule, $\;$ $\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C} = 2R$ $\;$ where $R$ is the circumradius of $\triangle ABC$
By cosine rule,
$\cos A = \dfrac{b^2 + c^2 - a^2}{2 b c}$ $\implies$ $b^2 + c^2 - a^2 = 2 b c \cos A$
$\cos B = \dfrac{c^2 + a^2 - b^2}{2 c a}$ $\implies$ $c^2 + a^2 - b^2 = 2 c a \cos B$
$\cos C = \dfrac{a^2 + b^2 - c^2}{2 a b}$ $\implies$ $a^2 + b^2 - c^2 = 2 a b \cos C$
Now,
$\begin{aligned}
\left(a^2 - b^2 + c^2\right) \tan B & = 2 c a \cos B \tan B \\\\
& = 2 c a \sin B \\\\
& = \dfrac{a b c}{R} \;\;\; \cdots \; (1a)
\end{aligned}$
$\begin{aligned}
\left(a^2 + b^2 - c^2\right) \tan C & = 2 a b \cos C \tan C \\\\
& = 2 a b \sin C \\\\
& = \dfrac{a b c}{R} \;\;\; \cdots \; (1b)
\end{aligned}$
$\therefore \;$ In view of equations $(1a)$ and $(1b)$ we have,
$\left(a^2 - b^2 + c^2\right) \tan B = \left(a^2 + b^2 - c^2\right) \tan C$
Hence proved.