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If the area of a triangle ABC is $Δ,$ then $a^{2} \sin 2 B + b^{2} \sin 2 A =$

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$2 Δ$

$4 Δ$

$3 Δ$

$- 4 Δ$

answer is C.

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Detailed Solution

$a^{2} \sin 2 B + b^{2} \sin 2 A = a^{2} .2 \sin B . \cos B + b^{2} .2 s i n A c o s A$
$= \frac{a^{2} b}{R} \cos B + b^{2} \frac{a}{R} \cos A = \frac{a b}{R} (a \cos B + b \cos A) = \frac{a b c}{R} = 4 Δ$

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