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Question
In Δ ABC, prove that, a (b cos C - c cos B) = b2 - c2.
Solution
Taking LHS
`a(bcosC-c cosB)`
`=abcosC-ac cosB`
`=ab((a^2+b^2-c^2)/(2ab))-ac((a^2+c^2-b^2)/(2ac))`
`(a^2+b^2-c^2-a^2-c^2+b^2)/2`
`=(2b^2-2c^2)/2`
`=b^2-c^2 "(RHS)"`
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