Answer :

Given: A + B + C = 0

To Prove:


Taking LHS,


Expanding along the first row, we get




= [1{1 – cos2A} – cos C {cos C – cos B cos A} + cos B {cos C cos A – cos B}]


= {1 – cos2A} – {cos2C – cos A cos B cos C} + {cos A cos B cos C – cos2B}


= {sin2A} – cos2C + cos A cos B cos C + cos A cos B cos C – cos2B


[ cos2x + sin2x = 1]


= sin2A – cos2C – cos2B + 2cos A cos B cos C


= -(cos2B – sin2A)– cos2C + 2cos A cos B cos C


= -[cos(B + A)cos(B – A)] + cos C [2 cos A cos B – cos C]


[ cos2B – sin2A = cos(B + A)cos(B – A)]


= -[cos(B + A)cos(B – A)] + cos C [cos(A + B) + cos(A – B) – cos C]…(i)


[ 2cos A cos B = cos(A + B) + cos(A – B)]


It is given that A + B + C = 0


A + B = - C


Putting the value of A + B in eq (i), we get


= -[cos(- C) cos(B – A)] + cos C [cos(-C) + cos (A – B) – cos C]


= -cos C cos(B – A) + cos C[cos C + cos(A – B) – cos C]


[ cos(-C) = cos C]


Now, cos(A – B) = cos A cos B + sin A sin B


= -cos C{cos B cos A + sin B sin A} + cos C [cos A cos B + sin A sin B]


= 0 = RHS


Hence Proved


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