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

Answer: Option B

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