If the eccentricity of a hyperbola is 3/2 then the angle between asymptotes of conjugate hyperbola is 
 

Given $\mathrm{e}=\frac{3}{2}$

$$\begin{gathered} we know that \\ \Rightarrow \frac{1}{{\mathrm{e}}^2}+\frac{1}{{{\mathrm{e}}_1}^2}=1 \\ \Rightarrow \frac{1}{{{\mathrm{e}}_1}^2}=1-\frac{4}{9} \\ \Rightarrow {{\mathrm{e}}_1}^2=\frac{9}{5} \\ \Rightarrow {\mathrm{e}}_1=\frac{3}{\sqrt{5}} \end{gathered}$$

let $\text{'}\mathrm{\theta }\text{'}$ be angle between asymptotes of conjugate hyperbola

$$\begin{gathered} \Rightarrow \mathrm{\theta }=2{\sec }^{-1}\left({\mathrm{e}}_1\right) \\ \Rightarrow sec\left(\frac{\theta }{2}\right)=e_1 \\ \cos \left(\frac{\mathrm{\theta }}{2}\right)=\frac{1}{{\mathrm{e}}_1} \\ \Rightarrow \cos \frac{\mathrm{\theta }}{2}=\frac{\sqrt{5}}{3} \\ \mathrm{Now} \mathrm{cos\theta }=2{\cos }^2\left(\frac{\mathrm{\theta }}{2}\right)-1 \\ =2\left(\frac{5}{9}\right)-1 \\ \mathrm{cos\theta }=\frac{1}{9} \\ \Rightarrow \mathrm{\theta }={\cos }^{-1}\left(\frac{1}{9}\right) \end{gathered}$$