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Question
In any Δ ABC, prove the following:
ac cos B - bc cos A = a2 - b2
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Solution
LHS = ac cos B - bc cos A = a2 - b2
`= "ac"(("c"^2 + "a"^2 - "b"^2)/"2ca") - "bc"(("b"^2 + "c"^2 - "a"^2)/"2bc")`
`= 1/2 ("c"^2 + "a"^2 - "b"^2) - 1/2 ("b"^2 + "c"^2 - "a"^2)`
`= 1/2 ("c"^2 + "a"^2 - "b"^2 - "b"^2 - "c"^2 + "a"^2)`
`= 1/2 (2"a"^2 - 2"b"^2)`
`= "a"^2 - "b"^2`
= RHS
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