If tan40° = α, then find $$\frac{{\tan {{320}^ \circ } - \tan {{310}^ \circ }}}{{1 + \tan {{320}^ \circ } \cdot \tan {{310}^ \circ }}}$$

Solution (By Examveda Team)

$$\eqalign{ & \frac{{\tan {{320}^ \circ } - \tan {{310}^ \circ }}}{{1 + \tan {{320}^ \circ }\tan {{310}^ \circ }}} \cr & = \frac{{\tan \left( {{{360}^ \circ } - {{40}^ \circ }} \right) - \tan \left( {{{270}^ \circ } + {{40}^ \circ }} \right)}}{{1 + \tan \left( {{{360}^ \circ } - {{40}^ \circ }} \right)\tan \left( {{{270}^ \circ } + {{40}^ \circ }} \right)}} \cr & = \frac{{ - \tan {{40}^ \circ } + \cot {{40}^ \circ }}}{{1 + \tan {{40}^ \circ } \times \cot {{40}^ \circ }}} \cr & = \frac{{\cot {{40}^ \circ } - \tan {{40}^ \circ }}}{{1 + 1}} \cr & = \frac{{\frac{1}{\alpha } - \alpha }}{2} \cr & = \frac{{1 - {\alpha ^2}}}{{2\alpha }} \cr} $$