State true or false
$(b^{2} + c^{2} - a^{2}) tan A = (c^{2} + a^{2} - b^{2}) tan B$ = $(a^{2} + b^{2} - c^{2}) tan C$ .

True

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False

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Solution

The correct option is A True
$(c^{2} - a^{2} + b^{2}) tan A = (C^{2} - a^{2} + b^{2}) \times \frac{sin A}{cos A}$

$= (c^{2} - a^{2} + b^{2}) \times \frac{a}{k \frac{b^{2} + c^{2} - a^{2}}{2 b c}}$

$(c^{2} - a^{2} + b^{2}) tan A = \frac{2 a b c}{k}$ ...(1)

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

$= (a^{2} - b^{2} + c^{2}) \times \frac{b}{k \frac{(a^{2} - b^{2} + c^{2})}{2 a c}}$

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

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

$(a^{2} + b^{2} - c^{2}) \frac{sin C}{cos C} = (a^{2} + b^{2} - c^{2}) \times c k \frac{(a^{2} + b^{2} - c^{2})}{2 a b}$

$= \frac{2 a b c}{k}$ (3)

from (1) (2) and (3)

$(c^{2} - a^{2} + b^{2}) tan A = (a^{-} b^{2} + c^{2}) tan B = (a^{2} + b^{2} - c^{2}) tan C$
Hence proved.

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