2cosec²23° cot²67° - sin²23° - sin²67° - cot²67° is equal to?

Solution(By Examveda Team)

According to the question,
  $$2{\operatorname{cosec} ^2}{23^ \circ }{\text{ co}}{{\text{t}}^2}{67^ \circ } - {\sin ^2}{23^ \circ } - $$       $${\sin ^2}{67^ \circ } - $$     $${\text{co}}{{\text{t}}^2}{67^ \circ }$$
  $$ \Rightarrow 2{\operatorname{cosec} ^2}{23^ \circ }{\text{ co}}{{\text{t}}^2}\left( {{{90}^ \circ } - {{23}^ \circ }} \right) - $$       $${\sin ^2}{23^ \circ } - $$   $${\sin ^2}\left( {{{90}^ \circ } - {{23}^ \circ }} \right) - $$     $${\text{co}}{{\text{t}}^2}{67^ \circ }$$
  $$ \Rightarrow 2{\operatorname{cosec} ^2}{23^ \circ }{\text{ ta}}{{\text{n}}^2}{23^ \circ } - $$     $$\left( {{{\sin }^2}{{23}^ \circ } + co{s^2}{{23}^ \circ }} \right) - $$   $${\text{co}}{{\text{t}}^2}{67^ \circ }$$
$$\eqalign{ & \Rightarrow \frac{2}{{{\text{co}}{{\text{s}}^2}{{23}^ \circ }}} - 1 - {\text{co}}{{\text{t}}^2}{67^ \circ } \cr & \Rightarrow 2se{c^2}{23^ \circ } - {\text{1}} - {\text{co}}{{\text{t}}^2}\left( {{{90}^ \circ } - {{23}^ \circ }} \right) \cr & \Rightarrow 2se{c^2}{23^ \circ } - {\text{1}} - {\text{ta}}{{\text{n}}^2}{23^ \circ } \cr & \Rightarrow 2se{c^2}{23^ \circ } - \left( {{\text{1}} + {\text{ta}}{{\text{n}}^2}{{23}^ \circ }} \right) \cr & \Rightarrow 2se{c^2}{23^ \circ } - {\sec ^2}{23^ \circ } \cr & \Rightarrow se{c^2}{23^ \circ } \cr} $$