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Question
If A + B + C = 180°, prove that `tan "A"/2 tan "B"/2 + tan "B"/2 tan "C"/2 + tan "C"/2 tan "A"/2` = 1
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Solution
Given A + B + C = 180°
⇒ `("A" + "" + "C")/2` = 90°
So `tan(("A" + "B")/2) = tan(90^circ - "C"/2) = cot "C"/2`
(i.e) `(tan "A"/2 + tan "B"/2)/(1 - tan "A"/2 tan "B"/2) = cot "C"/2 = 1/(tan "C"/2)`
⇒ `(tan "A"/2 + tan "B"/2)tan "C"/2 = 1 - tan "A"/2 tan "B"/2`
(i.e) `tan "A"/2 tan "C"/2 + tan "B"/2 tan "C"/2 = 1 - tan "A"/2 tan "B"/2`
(i.e) `tan "A"/2 tan "B"/2 + tan "B"/2 tan "C"/2 + tan "C"/2 tan "A"/2` = 1
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