Let $$\theta$$ be the acute angle between the tangents to the ellipse $$\frac{x^2}{9} + \frac{y^2}{1} = 1$$ and the circle $$x^2 + y^2 = 3$$ at their point of intersection in the first quadrant. Then $$\tan\theta$$ is equal to:

We have two curves, the ellipse $$\dfrac{x^{2}}{9}+\dfrac{y^{2}}{1}=1$$ and the circle $$x^{2}+y^{2}=3$$. First we locate their points of intersection in the first quadrant by solving the two equations simultaneously.

From the ellipse we get $$y^{2}=1-\dfrac{x^{2}}{9}.$$ Substituting this in the circle gives

$$x^{2}+1-\dfrac{x^{2}}{9}=3.$$

Collecting the $$x^{2}$$ terms,

$$x^{2}-\dfrac{x^{2}}{9}=3-1 \;\;\Longrightarrow\;\; \left(1-\dfrac{1}{9}\right)x^{2}=2.$$

Since $$1-\dfrac{1}{9}=\dfrac{8}{9},$$ we get

$$\dfrac{8}{9}x^{2}=2 \;\;\Longrightarrow\;\; x^{2}=2\cdot\dfrac{9}{8}=\dfrac{18}{8}=\dfrac{9}{4}.$$

Because we want the first-quadrant point, $$x=\dfrac{3}{2}>0.$$ Now we use $$y^{2}=1-\dfrac{x^{2}}{9}$$ to find $$y$$:

$$y^{2}=1-\dfrac{9/4}{9}=1-\dfrac{1}{4}=\dfrac{3}{4}\;\;\Longrightarrow\;\;y=\dfrac{\sqrt{3}}{2}>0.$$

Thus the common point is $$P\!\left(\dfrac{3}{2},\dfrac{\sqrt{3}}{2}\right).$$

Next we write the tangent to each curve at $$P$$.

For the ellipse $$\dfrac{x^{2}}{a^{2}}+\dfrac{y^{2}}{b^{2}}=1,$$ the tangent at $$(x_{1},y_{1})$$ is given by the standard formula $$\dfrac{xx_{1}}{a^{2}}+\dfrac{yy_{1}}{b^{2}}=1.$$ Here $$a^{2}=9,\;b^{2}=1,\;x_{1}=\dfrac{3}{2},\;y_{1}=\dfrac{\sqrt{3}}{2},$$ so

$$\dfrac{x\!\left(\dfrac{3}{2}\right)}{9}+y\!\left(\dfrac{\sqrt{3}}{2}\right)=1.$$

Simplifying, $$\dfrac{3x}{18}+\dfrac{\sqrt{3}}{2}y=1 \;\;\Longrightarrow\;\; \dfrac{x}{6}+\dfrac{\sqrt{3}}{2}y=1.$$

Solving for $$y$$ in slope-intercept form,

$$\dfrac{\sqrt{3}}{2}y=1-\dfrac{x}{6} \;\;\Longrightarrow\;\; y=\dfrac{2}{\sqrt{3}}\!\left(1-\dfrac{x}{6}\right)=\dfrac{2}{\sqrt{3}}-\dfrac{x}{3\sqrt{3}}.$$

Hence the slope of the ellipse tangent is

$$m_{e}=-\dfrac{1}{3\sqrt{3}}.$$

For a circle $$x^{2}+y^{2}=r^{2}$$ centred at the origin, the tangent at $$(x_{1},y_{1})$$ is $$xx_{1}+yy_{1}=r^{2}.$$ Using $$x_{1}=\dfrac{3}{2},\;y_{1}=\dfrac{\sqrt{3}}{2},\;r^{2}=3,$$ we obtain

$$x\!\left(\dfrac{3}{2}\right)+y\!\left(\dfrac{\sqrt{3}}{2}\right)=3.$$

Multiplying by $$2,$$ $$3x+\sqrt{3}\,y=6.$$ Solving for $$y,$$

$$\sqrt{3}\,y=6-3x \;\;\Longrightarrow\;\; y=\dfrac{6-3x}{\sqrt{3}}=2\sqrt{3}-x\sqrt{3}.$$

Thus the slope of the circle tangent is

$$m_{c}=-\sqrt{3}.$$

Now we need the acute angle $$\theta$$ between the two tangents. For lines with slopes $$m_{1}$$ and $$m_{2}$$ we use the formula

$$\tan\theta=\left|\dfrac{m_{2}-m_{1}}{1+m_{1}m_{2}}\right|.$$

Taking $$m_{1}=m_{e}=-\dfrac{1}{3\sqrt{3}}$$ and $$m_{2}=m_{c}=-\sqrt{3},$$

$$m_{2}-m_{1}=-\sqrt{3}-\left(-\dfrac{1}{3\sqrt{3}}\right) =-\sqrt{3}+\dfrac{1}{3\sqrt{3}} =\dfrac{-8}{3\sqrt{3}},$$

and

$$1+m_{1}m_{2}=1+\left(-\dfrac{1}{3\sqrt{3}}\right)\!\left(-\sqrt{3}\right)=1+\dfrac{1}{3}=\dfrac{4}{3}.$$

Therefore

$$\tan\theta=\left|\dfrac{-8/(3\sqrt{3})}{4/3}\right| =\dfrac{8}{3\sqrt{3}}\cdot\dfrac{3}{4} =\dfrac{2}{\sqrt{3}}.$$

Hence, the correct answer is Option C.