If A+ B+ C $= 180^{0}$ , then
sin 2 A+ sin 2 B -sin 2 C = cos A cos B sin C.

True

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False

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Solution

The correct option is B False
Given $A + B + C = 180^{\circ}$
$sin 2 A + sin 2 B - sin 2 C = sin 2 A + 2 cos (B + C) sin (B - C)$ $(∵ sin A - sin B = 2 cos \frac{A + B}{2} sin \frac{A - B}{2})$
$= 2 sin A cos A + 2 cos (180^{\circ} - A) sin (B - C)$ $(∵ sin 2 θ = 2 sin θ cos θ)$
$= 2 sin (180^{\circ} - (B + C)) cos A - 2 cos A sin (B - C)$ $(∵ cos (180^{\circ} - θ) = - cos θ)$
$= 2 cos A (sin (B + C) - sin (B - C))$
$= 4 cos A cos B sin C$ $(∵ sin (A + B) - sin (A - B) = 2 cos A sin B)$
So the given relation is $False$

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