Prove that `a(bcosC-c cosB)=b^2-c^2`

Explanation

`L.H.S. = a(bcosC - c cosB)`

From projection formula,

`a = bcosC+c cosB`

`:.a(bcosC - c cosB) = (bcosC+c cosB)(bcosC - c cosB)`

`=b^2cos^2C-c^2cos^2B`

`=b^2(1-sin^2C)-c^2(1-sin^2B)`

`=b^2-c^2 -(b^2sin^2C-c^2sin^2B)`

Now, from sine law,
`bsinC = csinB=>b^2sin^2C - c^2sin^2B = 0`

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