If $A, B, C$ are angles of triangle then
i) $\tan A + \tan B + \tan C = \tan A \tan B t an C$
ii) $\cot A \cot B + \cot B \cot C + \cot C \cot A = 1$

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Detailed Solution

i) $A + B + C = π \Rightarrow A + B = π - C$

Apply $‘ T a n ’$ on both sides

$Tan (A + B) = Tan (π - C)$

$\Rightarrow \frac{Tan A + Tan B}{1 - Tan A Tan B} = - Tan C$

$\Rightarrow Tan A + Tan B = - Tan C + Tan A Tan B Tan C$

$\Rightarrow Tan A + Tan B + Tan C = Tan A Tan B Tan C$

ii) $A + B + C = π \Rightarrow A + B = π - C$

Apply ‘Cot’ on both sides

$Cot (A + B) = Cot (π - C)$

$Cot A Cot B - 1 = - Cot C Cot B - Cot C \cot A$

$Cot A Cot B + Cot B Cot C + Cot C Cot A = 1$

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