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

True

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False

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Solution

The correct option is A True
In $△ A B C, A + B + C = 180^{\circ}$

$sin 2 A + sin 2 B - sin 2 C = sin 2 A + 2 cos (B + C) sin (B - C)$ $(∵ sin A - sin B = 2 cos \frac{A + B}{2} sin \frac{A - B}{2})$

$= 2 sin A cos A + 2 cos (180^{\circ} - A) sin (B - C)$ $(∵ sin 2 θ = 2 sin θ cos θ)$

$=2\sin (180^{\circ}-(B+C))\cos A-2\cos A\sin(B-

C)(\because \cos (180^{\circ}-\theta)=-\cos \theta)$

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

$= 4 cos A cos B sin C$ $(∵ sin (A + B) - sin (A - B) = 2 cos A sin B)$

So the given relation is $True$

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