The angle between the asymptotes of the hyperbola $x^2-3 y^2=3$ is

Given equation of hyperbola is $\begin{aligned} x^2-3 y^2 & =3 \\ \Rightarrow \quad \frac{x^2}{3}-\frac{y^2}{1} & =1 \end{aligned}$ Here, $a^2=3$ and $b^2=1, a>b$. Now, the equation of asymptote of this hyperbola is, $\begin{aligned} y & = \pm \frac{b}{a} x \\ \Rightarrow \quad y & = \pm \frac{1}{\sqrt{3}} x\end{aligned}$
Let slope of asymptote (i) is, $m_1=\frac{1}{\sqrt{3}}$ and slope of asymptote (ii) is, $m_2=\frac{-1}{\sqrt{3}}$ Let $\theta$ be the angle between both asymptote, then $\tan \theta=\left|\frac{m_1-m_2}{1+m_1 m_2}\right|=\left|\frac{1 / \sqrt{3}+1 / \sqrt{3}}{1-1 / 3}\right|$ $\begin{array}{cc}\Rightarrow & \tan \theta=\left|\frac{2 / \sqrt{3}}{2 / 3}\right|=\sqrt{3} \\ \Rightarrow & \tan \theta=\tan 60^{\circ} \\ \Rightarrow & \theta=\pi / 3\end{array}$