In ∆ABC, prove the following:

We know that

$\cos A=\frac{b^{2}+c^{2}-a^{2}}{2 b c}, \frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}=k$

So,

$\left(c^{2}-a^{2}+b^{2}\right) \tan A=\left(c^{2}-a^{2}+b^{2}\right) \frac{\sin A}{\cos A}$

$=\left(c^{2}-a^{2}+b^{2}\right) \sin A \frac{2 b c}{b^{2}+c^{2}-a^{2}}$

$=2 b c \sin A$

$=2 k a b c \quad \ldots(1)$

$\left(a^{2}-b^{2}+c^{2}\right) \tan B=\left(a^{2}-b^{2}+c^{2}\right) \frac{\sin B}{\cos B}$

$=\left(a^{2}-b^{2}+c^{2}\right) \sin B \frac{2 a c}{a^{2}+c^{2}-b^{2}}$

$=2 a c \sin B$

$=2 k a b c$             ...(2)

$\left(b^{2}-a^{2}+c^{2}\right) \tan C=\left(b^{2}-a^{2}+c^{2}\right) \frac{\sin C}{\cos C}$

$=\left(b^{2}-c^{2}+a^{2}\right) \sin C \frac{2 a b}{a^{2}+b^{2}-c^{2}}$

$=2 a b \sin C$

$=2 k a b c$     ....(3)

From (1), (2) and (3), we get:

$\left(c^{2}-a^{2}+b^{2}\right) \tan A=\left(a^{2}-b^{2}+c^{2}\right) \tan B=\left(b^{2}-c^{2}+a^{2}\right) \tan C$