If β is one of the angles between the normals to the ellipse, x^2+3...

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}