For the three internal angles of a triangle

A+B+C=\pi

\tan(A+B)=\tan(\pi−C)

Apply the sum identity for tangent function on both sides.

The tangent of the sum of two angles equals the sum of the tangents of the two angles divided by 1 minus the product of the tangents of the angles.

Considering \tan(-C) = -\tan C, we get

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

Since \tan\pi=0, the equation is simplified to

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

Multiply both sides by 1−\tan A\tan B

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

Hence

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