# In any triangle A

Let a, b, c be the sides of any triangle ABC. Then by applying the sine rule, we get  c = k sin C

Similarly, b = k sin B

And a = k sin A

Here we will consider RHS, so we get

RHS = (b2 – c2) sin A

Substituting corresponding values in the above equation, we get

= [( k sin B)2 - ( k sin C)2] sin A

= k2(sin2 B – sin2 C )sin A……….(ii)

But, Substituting the above values in equation (ii), we get

= k2(sin(B + C) sin(B - C)) sin A

But A + B + C = π B + C = π –A, so the above equation becomes,

= k2(sin(π –A) sin(B - C))sin A

But sin (π - θ) = sin θ

= k2(sin(A) sin(B - C))sin A

Rearranging the above equation we get

= (k sin(A))( sin(B - C))(k sin A)

From sine rule, a = k sin A, so the above equation becomes,

= a2 sin(B - C) = RHS

Hence proved

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