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For any triangle ABC, prove that a(bcos...

For any triangle ABC, prove that `a(bcosC-c cosB)=b^2-c^2`

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Given,
L.H.S.=`a(bcosC-c cosB) `
By, projection formula....(i)
`a=bcosC+c cosB`
Multiplying `(bcos C-c cosB)` on both sides of equation (i) we get
`a(bcosC-c cosB)=(bcosC+c cosB)(bcosC-c cosB) `
`=>a(bcosC-c cosB)=b^2cos^2C-c^2cos^2B `
`=>a(bcosC-c cosB)=b^2(1-sin^2C)-c^2(1-sin^2B) `
...
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