If $$\beta$$ is one of the angles between the normals to the ellipse, $$x^2 + 3y^2 = 9$$ at the points $$(3\cos\theta, \sqrt{3}\sin\theta)$$ and $$(-3\sin\theta, \sqrt{3}\cos\theta)$$; $$\theta \in (0, \frac{\pi}{2})$$; then $$\frac{2\cot\beta}{\sin 2\theta}$$ is equal to:

We have the ellipse $$x^2+3y^2=9$$.

For any curve written as $$F(x,y)=0$$, the gradient $$\left(\frac{\partial F}{\partial x},\frac{\partial F}{\partial y}\right)$$ gives a vector normal to the curve. A quicker route for conics is to use the well-known fact that the slope of the tangent at a point $$(x,y)$$ on the ellipse $$x^2+3y^2=9$$ is obtained by implicit differentiation:

$$\frac{d}{dx}\,(x^2)+\frac{d}{dx}\,(3y^2)=\frac{d}{dx}\,(9) \implies 2x+6y\frac{dy}{dx}=0.$$

Hence

$$\frac{dy}{dx}_{\text{tangent}}=-\frac{x}{3y}.$$

The normal is perpendicular to the tangent, so its slope is the negative reciprocal:

$$m_{\text{normal}}=\frac{3y}{x}.$$

Now we evaluate this slope at the two given points:

Point $$P(3\cos\theta,\;\sqrt3\sin\theta):$$

$$m_1=\frac{3(\sqrt3\sin\theta)}{3\cos\theta}= \sqrt3\,\tan\theta.$$

Point $$Q(-3\sin\theta,\;\sqrt3\cos\theta):$$

$$m_2=\frac{3(\sqrt3\cos\theta)}{-3\sin\theta}= -\sqrt3\,\cot\theta.$$

The angle $$\beta$$ between the two normals is found from the standard two-line formula

$$\tan\beta=\left|\frac{m_2-m_1}{1+m_1m_2}\right|.$$

Substituting $$m_1=\sqrt3\tan\theta$$ and $$m_2=-\sqrt3\cot\theta$$ we obtain

$$m_1m_2=\sqrt3\tan\theta\;(-\sqrt3\cot\theta)=-3,$$

$$m_2-m_1=-\sqrt3\cot\theta-\sqrt3\tan\theta=-\sqrt3(\cot\theta+\tan\theta).$$

Hence

$$\tan\beta=\left|\frac{-\sqrt3(\cot\theta+\tan\theta)}{1-3}\right|=\frac{\sqrt3}{2}\,(\tan\theta+\cot\theta).$$

Taking the reciprocal gives

$$\cot\beta=\frac{2}{\sqrt3(\tan\theta+\cot\theta)}.$$

We now simplify $$\tan\theta+\cot\theta$$:

$$\tan\theta+\cot\theta=\frac{\sin\theta}{\cos\theta}+\frac{\cos\theta}{\sin\theta} =\frac{\sin^2\theta+\cos^2\theta}{\sin\theta\cos\theta} =\frac{1}{\sin\theta\cos\theta}.$$

Therefore

$$\cot\beta=\frac{2}{\sqrt3}\;(\sin\theta\cos\theta).$$

We need the expression $$\dfrac{2\cot\beta}{\sin2\theta}$$. First note that $$\sin2\theta=2\sin\theta\cos\theta$$, so

$$2\cot\beta = 2\left(\frac{2\sin\theta\cos\theta}{\sqrt3}\right)=\frac{4\sin\theta\cos\theta}{\sqrt3}.$$

Hence

$$\frac{2\cot\beta}{\sin2\theta}= \frac{\dfrac{4\sin\theta\cos\theta}{\sqrt3}}{2\sin\theta\cos\theta} = \frac{4}{\sqrt3}\times\frac{1}{2}= \frac{2}{\sqrt3}.$$

This value is independent of $$\theta$$ (as long as $$\theta\in(0,\pi/2)$$), and it matches option B.

Hence, the correct answer is Option B.