Prove that $b^{2} sin 2 C + c^{2} sin 2 B = 2 b c sin A$ .

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Solution

$b^{2} sin 2 C + c^{2} sin 2 B$
$= b^{2} (2 sin C cos C) + c^{2} (2 sin B cos B)$
$= (2 b cos C) b sin C + (2 c cos B) c sin B$
$∵ \frac{b}{sin B} = \frac{c}{sin C} ∴ b sin C = c sin B$
$= 2 c sin B (b cos C + c cos B)$
$= 2 c sin B \times a$ where $b cos C + c cos B = a$
$= 2 a c . \frac{b sin A}{a}$
$∵ \frac{a}{sin A} = \frac{b}{sin B} ∴ a sin B = b sin A$
$= 2 a c \times \frac{b sin A}{a}$
$= 2 b c sin A$

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