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Question
With the usual notations, prove that `(sin("A" - "B"))/(sin ("A" + "B")) = ("a"^2 - "b"^2)/"c"^2`
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Solution
By the sine rule,
`"a"/"sin A" = "b"/"sin B" = "c"/"sin C" = "k"`
∴ a = k sin A, b = k sin B, c = k sin C
RHS = `("a"^2 - "b"^2)/"c"^2 = ("k"^2 "sin"^2 "A" - "k"^2 "sin"^2 "B")/("k"^2 "sin"^2 "C")`
`= ("sin"^2"A" - "sin"^2"B")/("sin"^2"C")`
`= ((sin "A" + sin "B")(sin "A" - sin "B"))/[sin{pi - ("A" + "B")}]^2` .....[∵ A + B + C = π]
`= (2 sin (("A" + "B")/2). cos (("A" - "B")/2) xx 2 cos (("A" + "B")/2). sin (("A" - "B")/2))/("sin"^2 ("A + B"))`
`= (2 sin (("A" + "B")/2). cos (("A" + "B")/2) xx 2 sin (("A" - "B")/2). cos (("A" - "B")/2))/("sin"^2 ("A + B"))`
`= ("sin" ("A" + "B"). sin("A" - "B"))/("sin"^2 ("A" + "B"))`
`= (sin ("A - B"))/(sin ("A + B"))` = LHS
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