Question
In a non right triangle \triangle ABC, show that
\tan A+\tan B+\tan C = \tan A\tan B\tan C
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