Question

Solve $\frac{\tan {160}^{\circ }-\tan {110}^{\circ }}{1+\tan {160}^{\circ }.\tan {110}^{\circ }}$

Answer 1

Given that:

$\frac{\tan {160}^{\circ }-\tan {110}^{\circ }}{1+\tan {110}^{\circ }\tan {160}^{\circ }}$

Since we know that $\tan (\theta -\varphi )=\frac{\tan \theta -\tan \varphi }{1+\tan \theta \tan \varphi }$

So

$\frac{\tan {160}^{\circ }-\tan {110}^{\circ }}{1+\tan {110}^{\circ }\tan {160}^{\circ }}=\tan ({160}^{\circ }-{110}^{\circ })$

$=\tan \left({50}^{\circ }\right)$Answer