$\frac{sin (A - C) + 2 sin A + sin (A + C)}{sin (B - C) + 2 sin B + sin (B + C)}$ is equal to-

tan A

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\frac{sin A}{sin B}

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\frac{cos A}{cos B}

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\frac{sin C}{cos B}

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Solution

The correct option is A $\frac{sin A}{sin B}$
The value of $\frac{sin (A - C) + 2 sin A + sin (A + C)}{sin (B + C) + 2 sin B + sin (B + C)}$
$= \frac{sin (A - C) + sin (A + C) + 2 sin A}{sin (B - C) + sin (B + C) + 2 sin B}$
$= \frac{2 sin (\frac{A - C + A + C}{2}) cos (\frac{A - C - A - C}{2}) + 2 sin A}{2 sin (\frac{B - C + B + C}{2}) cos (\frac{B - C - B - C}{2}) + 2 sin B}$
$= \frac{2 sin A cos C + 2 sin C}{2 sin B cos C + 2 sin B}$
$= \frac{2 sin A (cos C + 1)}{2 sin B (cos C + 1)} = \frac{sin A}{sin B}$

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