Question

In a non right triangle \triangle ABC, show that

\tan A+\tan B+\tan C = \tan A\tan B\tan C

Collected in the board: Trigonometry

Steven Zheng posted 4 months ago

Answer

In a triangle, the sum of three internal angles is equal to \pi

C = \pi - (A+B)

\tan C = \tan [\pi - (A+B)]=-\tan(A+B)

\tan C =- \dfrac{\tan A+\tan B}{1-\tan A\tan B}

\tan C(1-\tan A\tan B) = -\tan A-\tan B

Therefore,

\tan A+\tan B+\tan C = \tan A\tan B\tan C

Steven Zheng posted 4 months ago

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