If tan15° = 2 - 3 , then the value of tan15° cot75° + tan75° cot15° is?

Answer: Option A

$$\text{ tan 1}{5}^{\circ }\text{cot 7}{5}^{\circ }+\text{tan 7}{5}^{\circ }\text{cot 1}{5}^{\circ }$$
  $$=\text{ tan 1}{5}^{\circ }\text{cot }\left({90}^{\circ }-{15}^{\circ }\right)+$$      $$\text{tan}{\left({90}^{\circ }-15\right)}^{\circ }$$    $$\text{cot1}{5}^{\circ }$$
$$\begin{array}{rl}& =\text{ ta}{\text{n}}^2\text{1}{5}^{\circ }+\text{co}{\text{t}}^2\text{1}{5}^{\circ }\\ & =\text{ta}{\text{n}}^{2}15^{\circ }+\text{co}{\text{t}}^{2}15^{\circ }.....(i)\\ & \left[\mathbf{Formula}\right]\\ & \cot \left({90}^{\circ }-\theta \right)=\tan \theta \\ & \tan \left({90}^{\circ }-\theta \right)=\cot \theta \\ & \text{Put value of tan1}{5}^{\circ }\\ & \cot 15^{\circ }=\frac{1}{\text{tan1}{5}^{\circ }}\\ & \cot 15^{\circ }=\frac{1}{\left(2-\sqrt{3}\right)}\\ & \cot 15^{\circ }=\frac{1}{\left(2-\sqrt{3}\right)}\times \frac{\left(2+\sqrt{3}\right)}{\left(2+\sqrt{3}\right)}\\ & \cot 15^{\circ }=2+\sqrt{3}\\ & \text{Now put value in equation (i)}\\ & \text{ tan 1}{5}^{\circ }+\text{cot 1}{5}^{\circ }\\ & ={\left(2-\sqrt{3}\right)}^2+{\left(2+\sqrt{3}\right)}^2\\ & =4+3-4\sqrt{3}+4+3+4\sqrt{3}\\ & =14\end{array}$$