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Question
In ΔABC, A + B + C = π show that
`tan "A"/2 tan "B"/2 + tan "B"/2 tan "C"/2 + tan "C"/2tan "A"/2` = 1
Sum
Solution
In ΔABC, A + B + C = π
∴ `"A"/2 + "B"/2 + "C"/2 = pi/2`
∴ `"A"/2 + "B"/2 = pi/2 - "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) = 1/(tan("C"/2)`
∴ `tan("C"/2)*tan("A"/2) + tan("B"/2)*tan("C"/2) = 1 - tan("A"/2)*tan("B"/2)`
∴ `tan("A"/2)*tan("B"/2) + tan("B"/2)*tan("C"/2) + tan("C"/2)*tan("A"/2)` = 1.
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Factorization Formulae - Trigonometric Functions of Angles of a Triangle
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