In Delta ABC, asin (B - C) + bsin (C - A) + c | vedclass

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(A) Using the Sine Rule,we have $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = k.$So,$a = k \sin A, b = k \sin B, c = k \sin C.$Substitutin

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(A) Using the Sine Rule,we have $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = k.$So,$a = k \sin A, b = k \sin B, c = k \sin C.$Substituting these in the expression:$a \sin (B - C) + b \sin (C - A) + c \sin (A - B) = k [\sin A \sin (B - C) + \sin B \sin (C - A) + \sin C \sin (A - B)]$Since $A + B + C = \pi,$ we have $A = \pi - (B + C),$ so $\sin A = \sin (B + C).$Substituting $\sin A = \sin (B + C)$:$= k [\sin (B + C) \sin (B - C) + \sin (C + A) \sin (C - A) + \sin (A + B) \sin (A - B)]$Using the identity $\sin (x + y) \sin (x - y) = \sin^2 x - \sin^2 y$:$= k [(\sin^2 B - \sin^2 C) + (\sin^2 C - \sin^2 A) + (\sin^2 A - \sin^2 B)]$$= k [0] = 0.$

In $\Delta ABC,$ $a\sin (B - C) + b\sin (C - A) + c\sin (A - B) = $

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