Given \(A,B>0\) and \(A+B=\dfrac{\pi}{6}\). Using \[ \tan A+\tan B=\frac{\sin(A+B)}{\cos A\cos B}, \] with \(\sin(A+B)=\sin\frac{\pi}{6}=\dfrac12\). To minimize \(\tan A+\tan B\), maximize \(\cos A\cos B\) subject to \(A+B=\dfrac{\pi}{6}\).

The product \(\cos A\cos B\) (with fixed sum) is maximized at \(A=B=\dfrac{\pi}{12}\). Thus \[ \cos A\cos B\le \cos^2\!\frac{\pi}{12}=\frac{1+\cos\frac{\pi}{6}}{2} =\frac{1+\frac{\sqrt3}{2}}{2}=\frac{2+\sqrt3}{4}. \] Hence \[ \min(\tan A+\tan B)=\frac{\frac12}{\frac{2+\sqrt3}{4}} =\frac{2}{2+\sqrt3} =\boxed{\,4-2\sqrt3\,}. \]