Since, $x^2+3 y^2=9$
$$
\begin{aligned}
&\Rightarrow 2 x+6 y \frac{d y}{d x}=0 \\
&\Rightarrow \frac{d y}{d x}=\frac{-x}{3 y}
\end{aligned}
$$
Slope of normal is $-\frac{d x}{d y}=\frac{3 y}{x}$
$$
\begin{gathered}
\Rightarrow\left(-\frac{d y}{d x}\right)(3 \cos \theta, \sqrt{3} \sin \theta) \\
=\frac{3 \sqrt{3} \sin \theta}{3 \cos \theta}=\sqrt{3} \tan \theta=m_{\mathrm{l}} \\
\&\left(-\frac{d x}{d y}\right)(-3 \sin \theta, \sqrt{3} \cos \theta) \\
=\frac{3 \sqrt{3} \cos \theta}{-3 \sin \theta}=-\sqrt{3} \cot \theta=m_2
\end{gathered}
$$
As, $\beta$ is the anagle between the normals to the given ellipse then
$$
\begin{aligned}
&\tan \beta=\left|\frac{m_1-m_2}{1+m_1 m_2}\right| \\
&=\left|\frac{\sqrt{3} \tan \theta+\sqrt{3} \cot \theta}{1-3 \tan \theta \cot \theta}\right| \\
&=\left|\frac{\sqrt{3} \tan \theta+\sqrt{3} \cot \theta}{1-3}\right|
\end{aligned}
$$
So, $\tan \beta=\frac{\sqrt{3}}{2}|\tan \theta+\cot \theta|$
$$
\begin{aligned}
&\Rightarrow \quad \frac{1}{\cot \beta}=\frac{\sqrt{3}}{2}\left|\frac{\sin \theta}{\cos \theta}+\frac{\cos \theta}{\sin \theta}\right| \\
&\Rightarrow \quad \frac{1}{\cot \beta}=\frac{\sqrt{3}}{2}\left|\frac{1}{\sin \theta \cos \theta}\right| \\
&\Rightarrow \quad \frac{1}{\cot \beta}=\frac{\sqrt{3}}{\sin 2 \theta} \Rightarrow \frac{2 \cot \beta}{\sin 2 \theta}=\frac{2}{\sqrt{3}}
\end{aligned}
$$