$sin 2 A + sin 2 B - sin 2 C = 4 cos A cos B sin C$ .

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Solution

L.H.S $= sin 2 A + sin 2 B - sin 2 C$
$= 2 sin A cos A + 2 sin (B - C) cos (B + C)$
$= 2 sin A cos A - 2 sin (B - C) cos A$
$= 2 cos A [sin A - sin (B - C)]$
$= 2 cos A [sin (B + C) - sin (B - C)]$
$= 2 cos A [2 cos B sin C]$ $∵ sin A = sin (B + C)$
$= 4 cos A cos B sin C$ .

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