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With usual notatons , in triangle ABC , ...

With usual notatons , in `triangle ABC` , prove that `a(bcosC-c cos B)=b^2-c^2`.

Text Solution

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LHS `=a(bcosC-c cos B)`
`=a[b((a^2+b^2-c^2)/(2ab))-c((c^2+a^2-b^2)/(2ca))]`
`=((a^2+b^2-c^2)/(2))-((c^2+a^2-b^2)/(2))`
`=(a^2+b^2-c^2-c^2-a^2+b^2)/(2)`
`=(2b^2-2c^2)/(2)=b^2-c^2=RHS`.
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