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Question
In a ∆ABC, prove the following, `("a"^2 - "c"^2)/"b"^2 = (sin ("A" - "C"))/(sin("A" + "C"))`
Solution
`("a"^2 - "c"^2)/"b"^2 = (sin ("A" - "C"))/(sin("A" + "C"))`
We know sin(A + B) . sin(A – B) = sin2A – sin2B
Also `"a"/sin"B" = "b"/sin"B" = "c"/sin"C"` = 2R
`"a"/sin"A"` = 2R ⇒ a = 2R sin A
`"b"/sin"B"` = 2R ⇒ b = 2R sin B
`"c"/sin"C"` = 2R ⇒ c = 2R sin C
`(sin ("A" - "C"))/(sin("A" + "C")) = (sin("A" + "C")sin("A" - "C"))/(sin("A" + "C") sin("A" + "C"))`
= `(sin^2"A" - sin^2"C")/((sin(180^circ - "B")) sin(180^circ - "B"))`
= `(4"R"^2 (sin^2"A" - sin^2"C"))/(4"R"^2 sin"B" * sin "B")`
= `(4"R"^2 sin^2"A" - 4"R"^2 sin^2"C")/(4"R"^2 sin^2"B")`
= `((2"R" sin"A")^2 - (2"R" sin"C")^2)/(2"R" sin"B")^2`
`(sin ("A" - "C"))/(sin("A" + "C")) = ("a"^2 - "c"^2)/"b"^2`
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