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In ∆ABC, prove that ac cos B − bc cos A = a2 − b2

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Question

In ∆ABC, prove that ac cos B − bc cos A = a2 − b2 

Sum
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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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Chapter 1.3: Trigonometric Functions - Very Short Answers

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