Trig2.11

$A+B+C=180^\circ\,$ , prove that $\tan A+\tan B+\tan C=\tan A \tan B \tan C\,$

$A+B+C=180,A+B=180-C\,$

$\tan (A+B)=\tan (180-C)\,$

$\frac{\tan A+\tan B}{1-\tan A \tan B}=-\tan C\,$

Cross multiplying, we get

$\tan A+\tan B=-\tan C+\tan A \tan B \tan C\,$

$\tan A +\tan B+\tan C=\tan A \tan B \tan C\,$ which is the required.