If $$A > 0$$, $$B > 0$$ and $$A + B = \frac{\pi}{6}$$, then the minimum positive value of $$(\tan A + \tan B)$$ is:

We have $$A > 0,\; B > 0$$ and the fixed sum $$A + B = \dfrac{\pi}{6}\;.$$ Our task is to find the minimum positive value of $$\tan A + \tan B$$ under this condition.

We begin with the standard trigonometric identity

$$\tan X + \tan Y \;=\; \dfrac{\sin X}{\cos X} + \dfrac{\sin Y}{\cos Y} = \dfrac{\sin X \cos Y + \sin Y \cos X}{\cos X \cos Y} = \dfrac{\sin\,(X + Y)}{\cos X \cos Y}\;.$$

Applying this formula to the angles $$A$$ and $$B,$$ we obtain

$$\tan A + \tan B = \dfrac{\sin\,(A + B)}{\cos A \cos B}\;.$$

The numerator is easy to simplify because $$A + B = \dfrac{\pi}{6},$$ so

$$\sin\,(A + B) \;=\; \sin\!\left(\dfrac{\pi}{6}\right) = \dfrac12\;.$$

Thus

$$\tan A + \tan B = \dfrac{\dfrac12}{\cos A \cos B} = \dfrac{1}{2\,\cos A \cos B}\;.$$

Since the factor $$\dfrac12$$ in the numerator is a constant, minimizing $$\tan A + \tan B$$ is equivalent to maximizing the product $$\cos A \cos B.$$

Because $$B = \dfrac{\pi}{6} - A,$$ we can view the product solely as a function of $$A$$:

$$f(A) \;=\; \cos A \;\cos\!\Bigl(\dfrac{\pi}{6} - A\Bigr),$$, $$(0 < A < \dfrac{\pi}{6}).$$

To locate the maximum of $$f(A)$$ we differentiate and set the derivative to zero. First compute the derivative:

$$ \begin{aligned} f'(A) &= -\sin A \;\cos\!\Bigl(\dfrac{\pi}{6} - A\Bigr) \;+\; \cos A \;\sin\!\Bigl(\dfrac{\pi}{6} - A\Bigr)\;. \end{aligned} $$

Setting $$f'(A)=0$$ gives

$$\cos A \;\sin\!\Bigl(\dfrac{\pi}{6} - A\Bigr) = \sin A \;\cos\!\Bigl(\dfrac{\pi}{6} - A\Bigr).$$

Dividing both sides by $$\cos A\;\cos\!\bigl(\dfrac{\pi}{6} - A\bigr)$$ (which is positive in the interval) yields

$$\tan\!\Bigl(\dfrac{\pi}{6} - A\Bigr) \;=\; \tan A\;.$$

The equality of the tangents forces their arguments to be equal (modulo $$\pi$$, but only the principal value lies inside our interval), so

$$\dfrac{\pi}{6} - A \;=\; A$$ $$\Longrightarrow$$ $$2A \;=\; \dfrac{\pi}{6}$$ $$\Longrightarrow$$ $$A \;=\; B \;=\; \dfrac{\pi}{12}\;.$$

Thus the product $$\cos A \cos B$$ attains its maximum when both angles are equal to $$\dfrac{\pi}{12}.$$ Let us now evaluate that product explicitly:

$$ \cos A \cos B = \cos^2\!\Bigl(\dfrac{\pi}{12}\Bigr). $$

Recall the exact value of $$\cos\dfrac{\pi}{12}$$ (which is $$\cos15^\circ$$):

$$\cos\dfrac{\pi}{12} = \cos 15^\circ = \dfrac{\sqrt6 + \sqrt2}{4} = \dfrac{\sqrt{2 + \sqrt3}}{2}\;.$$

Either form may be used; squaring the simpler radical form gives

$$ \cos^2\!\Bigl(\dfrac{\pi}{12}\Bigr) = \left(\dfrac{\sqrt{2 + \sqrt3}}{2}\right)^2 = \dfrac{\,2 + \sqrt3\,}{4}\;. $$

Now substitute this maximum product into the earlier expression for $$\tan A + \tan B$$:

$$ \tan A + \tan B\Bigl|_{\min} = \dfrac{1}{2 \,\cos A \cos B} = \dfrac{1}{2 \,\dfrac{\,2 + \sqrt3\,}{4}} = \dfrac{1}{\dfrac{\,2 + \sqrt3\,}{2}} = \dfrac{2}{\,2 + \sqrt3}\;. $$

Finally, rationalise the denominator:

$$ \dfrac{2}{\,2 + \sqrt3}\; \cdot\; \dfrac{\,2 - \sqrt3\,}{\,2 - \sqrt3\,} = \dfrac{2(2 - \sqrt3)}{(2 + \sqrt3)(2 - \sqrt3)} = \dfrac{\,4 - 2\sqrt3\,}{\,4 - 3\,} = 4 - 2\sqrt3\;. $$

Therefore the minimum positive value of $$(\tan A + \tan B)$$ is $$4 - 2\sqrt3,$$ which matches Option B.

Hence, the correct answer is Option B.