Question -
Answer -
Upon simplification we get,
= k2 [sin A sin (B – C) + sin B sin (C– A) + sin C sin (A – B)]
We know, sin (A – B) = sin A cos B – cos A sin B
Sin (B – C) = sin B cos C – cos B sin C
Sin (C – A) = sin C cos A – cos C sin A
So the above equation becomes,
= k2 [sin A (sin B cos C – cos B sinC) + sin B (sin C cos A – cos C sin A) + sin C (sin A cos B – cos A sin B)]
= k2 [sin A sin B cos C – sin A cos Bsin C + sin B sin C cos A – sin B cos C sin A + sin C sin A cos B – sin C cos Asin B)]
Upon simplification we get,
= 0
= RHS
Hence proved.