If the angle between asymptotes of a hyperbola is $\frac{π}{6}$ then the eccentricity of the hyperbola is

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$\sqrt{6} + \sqrt{2}$

$\begin{array}{ll} \sqrt{6} + \sqrt{3} \end{array}$

$2 + \sqrt{3}$

$\sqrt{5} - \sqrt{3}$

answer is A.

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Detailed Solution

Angle between two asymptotes are $2 \tan^{- 1} (\frac{b}{a})$ and $π - 2 \tan^{- 1} (\frac{b}{a})$

Let $\frac{π}{6} = 2 \tan^{- 1} (\frac{b}{a})$

$\begin{array}{l} \Rightarrow \frac{b}{a} = \tan (\frac{π}{12}) = \tan (\frac{π}{3} - \frac{π}{4}) = \frac{\sqrt{3} - 1}{\sqrt{3} + 1} = 2 - \sqrt{3} \\ \Rightarrow e^{2} - 1 = {(\frac{b}{a})}^{2} = 4 + 3 - 4 \sqrt{3} \\ \Rightarrow e^{2} = 8 - 4 \sqrt{3} = (\sqrt{6} - \sqrt{2})^{2} \\ \Rightarrow e = \sqrt{6} - \sqrt{2} \end{array}$

If $\frac{π}{6} = π - 2 \tan^{- 1} (\frac{b}{a})$

$\begin{array}{l} \Rightarrow \tan^{- 1} (\frac{b}{a}) = \frac{π}{2} - \frac{π}{12} \\ \Rightarrow \frac{b}{a} = \cot (\frac{π}{12}) = 2 + \sqrt{3} \\ \Rightarrow e = \sqrt{6} + \sqrt{2} \end{array}$