In ∆ABC, prove the following:
$(c^{2} - a^{2} + b^{2}) \tan A = (a^{2} - b^{2} + c^{2}) \tan B = (b^{2} - c^{2} + a^{2}) \tan C$

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Solution

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

$So, (c^{2} - a^{2} + b^{2}) \tan A = (c^{2} - a^{2} + b^{2}) \frac{\sin A}{\cos A} = (c^{2} - a^{2} + b^{2}) \sin A \frac{2 b c}{b^{2} + c^{2} - a^{2}} = 2 b c \sin A = 2 k a b c . . . (1) (a^{2} - b^{2} + c^{2}) \tan B = (a^{2} - b^{2} + c^{2}) \frac{\sin B}{\cos B} = (a^{2} - b^{2} + c^{2}) \sin B \frac{2 a c}{a^{2} + c^{2} - b^{2}} = 2 a c \sin B = 2 k a b c . . . (2) (b^{2} - a^{2} + c^{2}) \tan C = (b^{2} - a^{2} + c^{2}) \frac{\sin C}{\cos C} = (b^{2} - c^{2} + a^{2}) \sin C \frac{2 a b}{a^{2} + b^{2} - c^{2}} = 2 a b \sin C = 2 k a b c . . . . (3)$

From (1), (2) and (3), we get:
$(c^{2} - a^{2} + b^{2}) \tan A = (a^{2} - b^{2} + c^{2}) \tan B = (b^{2} - c^{2} + a^{2}) \tan C$

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Similar questions

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

In any $Δ A B C$ , prove that

$\frac{(b^{2} - c^{2})}{a^{2}} s i n 2 A + \frac{(c^{2} - a^{2})}{b^{2}} s i n 2 B + \frac{(a^{2} - b^{2})}{c^{2}} s i n 2 C = 0$

a, b, c

G . P .

x, y, z

G . P .

x = (b^{2} - c^{2}) \frac{tan B + tan C}{tan B - tan C}

y = (c^{2} - a^{2}) \frac{tan C + tan A}{tan C - tan A}

z = (a^{2} - b^{2}) \frac{tan A + tan B}{tan A - tan B}

a \sin \frac{A}{2} \sin (\frac{B - C}{2}) + b \sin \frac{B}{2} \sin (\frac{C - A}{2}) + c \sin \frac{C}{2} \sin (\frac{A - B}{2}) = 0

Δ A B C

c o s A = \frac{b^{2} + c^{2} - a^{2}}{2 b c}, c o s B = \frac{c^{2} + a^{2} - b^{2}}{2 a c}, c o s C = \frac{a^{2} + b^{2} - c^{2}}{2 a b}

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