$(c^{2} - a^{2} + b^{2}) t a n A = (a^{2} - b^{2} + c^{2}) t a n B = (b^{2} - c^{2} + a^{2}) t a n C$

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Solution

$For any Δ A B C, we have c o s A = \frac{b^{2} + c^{2} - a^{2}}{2 b c} c o s B = \frac{a^{2} + c^{2} - b^{2}}{2 a c} c o s C = \frac{a^{2} + b^{2} - c^{2}}{2 a b} Therefore, (c^{2} - b^{2} - a^{2}) t a n A = (c^{2} + b^{2} - a^{2}) \frac{s i n A}{c o s A} = (c^{2} + b^{2} - a^{2}) \frac{k a}{\frac{b^{2} + c^{2} - a^{2}}{2 b c}} = 2 k a b c Also, (a^{2} + c^{2} - b^{2}) t a n B = (a^{2} + c^{2} - b^{2}) \frac{s i n B}{c o s B} = (a^{2} + c^{2} - b^{2}) \frac{k b}{\frac{a^{2} + c^{2} - b^{2}}{2 a c}} Now, = 2 k a b c (a^{2} + b^{2} - c^{2}) t a n C = (a^{2} + b^{2} - c^{2}) \frac{s i n C}{c o s C} = (a^{2} + b^{2} - c^{2}) \frac{k c}{\frac{a^{2} + b^{2} - c^{2}}{2 a b}} = 2 k a b c Hence proved.$

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