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Prove that : \(a\,cos\,(\frac{B-C}{2})=(b\,+\,c)\,sin\frac{A}{2}.\) |
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Answer» LHS = \(a\,cos\frac{B-C}{2}\) \(=k\,sin\,A\,cos\frac{B-C}{2}\) \(=k\,sin\,(B+C)\,cos\frac{B-C}{2}\) [ ∴ A + B +C = 180° ∴ sin A = sin [180° − (B + C) ] \(=\frac{cos\frac{C}{2}cos\frac{A-B}{2}}{sin\frac{C}{2}cos\frac{C}{2}}\) \(=\frac{cos\frac{A-B}{2}}{sin\frac{C}{2}}\) = RHS ∴ LHS = RHS |
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