Answer :

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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