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