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Q.

If $A + B + C = 180 °$ , then $\sin 2 A + \sin 2 B - \sin 2 C =$

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a

$4 \cos A \cos B \sin C$

b

$4 \cos A \sin B \cos C$

c

$4 \sin A \cos B \sin C$

d

$4 \cos A \sin B \sin C$

answer is C.

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Detailed Solution

$\sin 2 A + \sin 2 B - \sin 2 C$

$2 \sin (A + B) \cos (A - B) - 2 \sin C$

$2 \sin C \cos (A - B) - 2 \sin C \cos C$

$2 \sin C (\cos (A - B) - \cos C)$

$2 \sin C (\cos (A - B) + \cos (A + B))$

$2 \sin C (2 \cos A \cos B)$

$4 \cos A \cos B \sin C$

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