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Question
In a ∆ABC, prove the following, `("a"+ "b")/("a" - "b") = tan(("A" + "B")/2) cot(("A" - "B")/2)`
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Solution
We know `"a"/sin"A" = "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
`("a" +"b")/("a" - "b") = (2"R" sin"A" + 2"R" sin"B")/(2"R" sin"A" - 2"R" sin"B")`
= `(2"R" (sin "A" + sin "B"))/(2"R" (sin"A" - sin "B"))`
= `(sin "A" + sin "B")/(sin "A" - sin "B")`
= `(2sin(("A" + "B")/2) * cos(("A" - "B")/2))/(2cos(("A" + "B")/2) * sin(("A" - "B")/2))`
= `(sin(("A" + "B")/2))/(cos(("A" + "B")/2)) xx (cos(("A" - "B")/2))/(sin(("A" - "B")/2))`
`("a"+ "b")/("a" - "b") = tan(("A" + "B")/2) * cot(("A" - "B")/2)`
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