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B
a2+b2c2
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C
c2a2−b2
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D
c2a2+b2
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Solution
The correct option is Aa2−b2c2 sin(A−B)sin(A+B=sinAcosB−sinBcosAsinC =accosB−bccosA
But cos B=a2+c2−b22ac,cosA=b2+c2−a22bc ⇒accosB−bccosA=12c2 (a2+c2−b2−b2−c2+a2)=a2−b2c2