Question

$\tan {40}^{\circ }+\tan {60}^{\circ }+\tan {80}^{\circ }=\tan {40}^{\circ }.\tan {60}^{\circ }.\tan {80}^{\circ }$

Answer 1

$\begin{array}{c}\tan {40}^{\circ }+\tan {60}^{\circ }+\tan {80}^{\circ }=\tan 40+\tan 60+\tan \left(180-100\right)\\ =\tan 40+\tan 60-\tan 100\\ =\tan 40+\tan 60-\frac{\tan 40+\tan 60}{1-\tan 40+\tan 60}\\ =\tan 40+\tan 60\left[1-\frac{1}{1-\tan 40\tan 60}\right]\\ =-\tan 40+\tan 60\left[\frac{tan40\tan 60}{1-\tan 40\tan 60}\right]\\ =-\tan 40\tan 60\tan 80\\ =\tan 40\tan 60\tan 80\end{array}$