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