Question

Prove that:

$\tan {60}^{\circ }\tan {80}^{\circ }\tan {20}^{\circ }=\tan {80}^{\circ }-\tan {20}^{\circ }-\tan {60}^{\circ }$

Answer 1

$\tan {60}^{\circ }=\tan ({80}^{\circ }-{60}^{\circ })$

$\tan {60}^{\circ }=\frac{\tan {80}^{\circ }-\tan {20}^{\circ }}{1+\tan {80}^{\circ }\cdot \tan {20}^{\circ }}$
$\tan {60}^{\circ }+\tan {80}^{\circ }.\tan {60}^{\circ }.\cdot \tan {20}^{\circ }\cdot \tan {60}^{\circ }=\tan {80}^{\circ }-\tan {20}^{\circ }$
$\tan {80}^{\circ }\cdot \tan {20}^{\circ }\cdot \tan {60}^{\circ }=\tan {80}^{\circ }-\tan {20}^{\circ }-\tan {60}^{\circ }$