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Question -

a (sin B тАУ sin C) + b (sin C тАУ sin A) + c (sin A тАУ sin B) = 0



Answer -

By using the sine rule we know,

a = k sin A, b = k sin B, c = k sin C

Let us consider LHS:

a (sin B тАУ sin C) + b (sin C тАУ sin A) + c (sin A тАУ sin B)

Substituting the values of a, b, c from sine rule inabove equation, we get

a (sin B тАУ sin C) + b (sin C тАУ sin A) + c (sin A тАУ sinB) = k sin A (sin B тАУ sin C) + k sin B (sin C тАУ sin A) + k sin C (sin A тАУ sin B)

= k sin A sin B тАУ k sin A sin C + k sin B sin C тАУ ksin B sin A + k sin C sin A тАУ k sin C sin B

Upon simplification, we get

= 0

= RHS

Hence proved.

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