2 sin 68^ cos 2 2^ - 2 cot 15^ 5 75^ - 3 45^ . 20^ . 40^ . 50^ . 70^ 5 is equal to?

Answer: Option C

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