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CGP EDU Academic Team
Published on: August 13, 2026
Prove that a 3 cos (B – C) + b 3 cos (C – A) + c 3 cos (A – B) = 3abc
Text Solution
Verified by ExpertsThe correct answer is:
A
Sol. a 3 cos(B – C) = a 2 . 2R sin Acos (B – C)
( sin (B + C) = sin A = 2Ra 2 sin (B + C) cos (B – C)
= Ra 2 [sin 2B + sin 2C]
= 2Ra 2 [sin B cos B + sin C cos C]
= a 2 [bcos B + C cos C]
Similarly b 3 cos (C –A) = b 2 [C cos C + a cos A]
C 3 cos (A – B) = C 2 [a cos A + b cos B]
L.H.S. = (a 2 b cos B + ab 2 cos A ) + (b 2 C cosC + bC 2 cosB) + C 2 a cos A + a 2 C cos C)
= ab + bc(a) + ca(b) = 3abc = RHS
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