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

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Solution

Consider:
$(c^{2} + b^{2} - a^{2}) \tan A = (c^{2} + b^{2} - a^{2}) \sin A \frac{2 b c}{(c^{2} + b^{2} - a^{2})} = 2 b c \sin A = 2 k a b c (∵ \frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{c} = k)$ ....(1)
Similarly,

$(a^{2} + c^{2} - b^{2}) \tan B = (a^{2} + c^{2} - b^{2}) \sin B \frac{2 a c}{(a^{2} + c^{2} - b^{2})} = 2 a c \sin B = 2 k a b c$ ...(2)
and
$(a^{2} + b^{2} - c^{2}) \tan C = (a^{2} + b^{2} - c^{2}) \sin C \frac{2 a b}{(a^{2} + b^{2} - c^{2})} = 2 a b \sin C = 2 k a b c$ ...(3)

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

Hence proved.

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Similar questions

(b^{2} + c^{2} - a^{2}) tan A = (c^{2} + a^{2} - b^{2}) tan B

(a^{2} + b^{2} - c^{2}) tan 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 \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}

In any $Δ A B C,$ prove that

$(b^{2} - c^{2}) c o t A + (c^{2} - a^{2}) c o t B + (A^{2} - B^{2}) c o t C = 0$

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