$$\frac{{2{\text{sin }}{{68}^ \circ }}}{{{\text{cos 2}}{{\text{2}}^ \circ }}}$$   $$ - $$ $$\frac{{2{\text{cot 1}}{5^ \circ }}}{{5\tan {{75}^ \circ }}}$$   $$ - $$ $$\frac{{3\tan {{45}^ \circ }.\tan {{20}^ \circ }.\tan {{40}^ \circ }.\tan {{50}^ \circ }.\tan {{70}^ \circ }}}{5}$$         is equal to?

Solution(By Examveda Team)

  $$\frac{{2{\text{sin }}{{68}^ \circ }}}{{{\text{cos 2}}{{\text{2}}^ \circ }}}$$   $$ - $$ $$\frac{{2{\text{cot 1}}{5^ \circ }}}{{5\tan {{75}^ \circ }}}$$   $$ - $$ $$\frac{{3\tan {{45}^ \circ }.\tan {{20}^ \circ }.\tan {{40}^ \circ }.\tan {{50}^ \circ }.\tan {{70}^ \circ }}}{5}$$
  $$ \Rightarrow \frac{{2{\text{sin }}{{68}^ \circ }}}{{{\text{cos }}\left( {{{90}^ \circ } - {{68}^ \circ }} \right)}} - $$     $$\frac{{2{\text{cot 1}}{5^ \circ }}}{{5\tan \left( {{{90}^ \circ } - {{15}^ \circ }} \right)}} - $$     $$\frac{{3.1\left( {\tan {{20}^ \circ }.\tan {{70}^ \circ }} \right).\left( {\tan {{40}^ \circ }.\tan {{50}^ \circ }} \right)}}{5}$$
$$\eqalign{ & \Rightarrow \frac{{2{\text{sin }}{{68}^ \circ }}}{{{\text{sin 6}}{{\text{8}}^ \circ }}} - \frac{{2{\text{cot 1}}{5^ \circ }}}{{5\cot {{15}^ \circ }}} - \frac{{3 \times 1 \times 1 \times 1}}{5} \cr & \left[ {{\text{If tan A}}{\text{.tan B}} = {\text{1 then, A}} + {\text{B}} = {{90}^ \circ }} \right] \cr & \Rightarrow 2 - \frac{2}{5} - \frac{3}{5} \cr & \Rightarrow 1 \cr} $$