The value of $$\frac{{2\tan {{60}^ \circ }}}{{1 + {{\tan }^2}{{60}^ \circ }}} = ?$$

Solution (By Examveda Team)

$$\eqalign{ & \frac{{2\tan {{60}^ \circ }}}{{1 + {{\tan }^2}{{60}^ \circ }}} \cr & = \frac{{2 \times \sqrt 3 }}{{1 + 3}} \cr & = \frac{{\sqrt 3 }}{2} \cr & = \sin {60^ \circ } \cr} $$