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Question
In ∆ABC, prove that ac cos B − bc cos A = a2 − b2
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Solution
ac cos B – bc cos A
= `"ac" (("c"^2 + "a"^2 - "b"^2)/(2"ac")) - "bc" (("b"^2 + "c"^2 - "a"^2)/(2"bc"))` .......[By cosine rule]
= `(("c"^2 + "a"^2 - "b"^2)/2) - (("b"^2 + "c"^2 - "a"^2)/2)`
= `("c"^2 + "a"^2 - "b"^2 - "b"^2 - "c"^2 + "a"^2)/2`
= `(2"a"^2 - 2"b"^2)/2`
= a2 – b2
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