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

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

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

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

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

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