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Question
In ΔABC, A + B + C = π show that
`cot "A"/2 + cot "B"/2 + cot "C"/2 = cot "A"/2 cot "B"/2 cot "C"/2`
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Solution
In ΔABC,
A + B + C = π
∴ A + B = π – C
∴ `tan(("A" + "B")/2) = tan((pi - "C")/2)`
∴ `tan("A"/2 + "B"/2) = tan(pi/2 - "C"/2)`
∴ `(tan "A"/2 + tan "B"/2)/(1 - tan "A"/2*tan "B"/2) = cot "C"/2`
∴ `(tan "A"/2 + tan "B"/2)/(1 - tan "A"/2*tan "B"/2) = 1/(tan "C"/2)`
∴ `tan "C"/2*(tan "A"/2 + tan "B"/2) = 1 - tan "A"/2*tan "B"/2`
∴ `tan "B"/2*tan "C"/2 + tan "A"/2*tan "C"/2 + tan "A"/2*tan "B"/2` = 1
Dividing throughout by `tan "A"/2*tan "B"/2*tan "C"/2`, we get
`1/(tan "A"/2) + 1/(tan "B"/2) + 1/(tan "C"/2) = 1/(tan "A"/2*tan "B"/2*tan "C"/2)`
∴ `cot "A"/2 + cot "B"/2 + cot "C"/2 = cot "A"/2 cot "B"/2 cot "C"/2`
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