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Question
In any Δ ABC, prove the following:
`("b" - "c")/"a" = (tan "B"/2 - tan "C"/2)/(tan "B"/2 +tan "C"/2)`
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Solution
By sine rule,
`"a"/"sin A" = "b"/"sin B" = "c"/"sin C"` = k
∴ a = k sin A, b = k sin B, c = k sin C
LHS = `("b" - "c")/"a"`
`= ("k sin B - k sin C")/"k sin A"`
`= ("sin B - sin C")/"sin A"`
`= ("sin B - sin C")/(sin {pi - ("B" + "C")}) ....[because "A + B + C" = pi]`
`= ("sin B - sin C")/(sin ("B + C"))`
`= (2 cos (("B + C")/2). sin (("B" - "C")/2))/(2 sin (("B + C")/2). cos (("B" + "C")/2))`
`= (sin ("B - C")/2)/(sin ("B" + "C")/2)`
`= sin("B"/2 - "C"/2)/sin ("B"/2 + "C"/2)`
`= (sin "B"/2 cos "C"/2 - cos "B"/2 sin "C"/2)/(sin "B"/2 cos "C"/2 + cos "B"/2 sin "C"/2)`
`= ((sin "B"/2 cos "C"/2)/(cos "B"/2 cos "C"/2) - (cos "B"/2 sin "C"/2)/(cos "B"/2 cos "C"/2))/((sin "B"/2 cos "C"/2)/(cos "B"/2 cos "C"/2) + (cos "B"/2 sin "C"/2)/(cos "B"/2 cos "C"/2))`
`= ((sin "B"/2)/(cos "B"/2) - (sin "C"/2)/(cos "C"/2))/((sin "B"/2)/(cos "C"/2) + (sin "C"/2)/(cos "C"/2))`
`= (tan "B"/2 - tan "C"/2)/(tan "B"/2 + tan "C"/2)`
= RHS.
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