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Miscellaneous Exercise 3 · Q105

Q.In △ABC\triangle ABC prove that a2sin⁡(B−C)=(b2−c2)sin⁡Aa^2\sin(B-C) = (b^2-c^2)\sin A.

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R.H.S. =(b2−c2)sin⁡A=k2(sin⁡2B−sin⁡2C)sin⁡A=k2sin⁡(B+C)sin⁡(B−C)sin⁡A=(b^2-c^2)\sin A=k^2(\sin^2B-\sin^2C)\sin A=k^2\sin(B+C)\sin(B-C)\sin A (using the same product identity as II.11A). Since B+C=π−A  ⟹  sin⁡(B+C)=sin⁡AB+C=\pi-A\implies\sin(B+C)=\sin A: R.H.S. $=k^2\sin^2A\sin(B-C)=(k\sin …

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