In a triangle ABC, $(a^{2} - b^{2} - c^{2}) tan A + (a^{2} - b^{2} + c^{2}) tan B$ is equal to :

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

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

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

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none of these

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Solution

The correct option is D none of these
$tan A = \frac{sin A}{cos A} = \frac{2 b c sin A}{b^{2} + c^{2} - a^{2}}$
$⟹ (a^{2} - b^{2} - c^{2}) tan A = - 2 b c sin A$
Similarly $(a^{2} + c^{2} - b^{2}) tan B = 2 a c sin B$
$(a^{2} - b^{2} - c^{2}) tan A + (a^{2} - b^{2} + c^{2}) tan B = - 2 b c sin A + 2 c a sin B = - 2 b (c sin A) + 2 a (c sin B) = - 2 b a sin C + 2 a b sin C = 0$

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Similar questions

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

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

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

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

\frac{a^{2} + b^{2} + c^{2}}{R^{2}}

8

a^{2} = b^{2} + c^{2}

b^{2} = c^{2} + a^{2}

c^{2} = a^{2} + b^{2}

In any $△ A B C, \frac{1}{r_{1}^{2}} + \frac{1}{r_{2}^{2}} + \frac{1}{r_{3}^{2}} + \frac{1}{r^{2}}$ is equal to

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