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In ΔABC, A + B + C = π show that cot A2+cot B2+cot C2=cot A2 cot B2cot C2

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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`

Sum
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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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Chapter 3: Trigonometry - 2 - Exercise 3.5 [Page 54]

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