Question

if tan abc are angles of triangle and none of them equal to π/2 then prove that tanA+ tanB+tanC=tanA.tanB.tanC
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Answer 1

SOLUTION

GIVEN

A, B, C are angles of triangle and none of them equal to π/2

TO PROVE

tan A + tan B + tan C = tan A . tanB . tanC

FORMULA TO BE IMPLEMENTED

$\displaystyle \sf{ \tan(A + B) = \frac{ \tan A + \tan B}{1 - \tan A \tan B} }$

EVALUATION

Here it is given that

$\displaystyle \sf{A + B + C = \pi}$

$\displaystyle \sf{ \implies \: A + B = \pi - C}$

$\displaystyle \sf{ \implies \tan(A + B) = \tan( \pi - C)}$

$\displaystyle \sf{ \implies \tan(A + B) = - \tan C}$

$\displaystyle \sf{ \implies \frac{ \tan A + \tan B}{1 - \tan A \tan B} = - \tan C}$

$\displaystyle \sf{ \implies \tan A + \tan B = \: - \tan C + \tan A \tan B \tan C}$

$\displaystyle \sf{ \implies \tan A + \tan B + \tan C = \tan A \tan B \tan C}$

Hence proved

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