Given

$A>0,\, B>0,\, A+B=\frac{\pi}{6}$ We need the minimum value of $\tan A+\tan B.$

Use the identity for $\tan(A+B)$

$\tan(A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}$

Since $A+B=\frac{\pi}{6}$, $\tan\left(\frac{\pi}{6}\right)=\frac{1}{\sqrt{3}}.$ Therefore, $\frac{\tan A+\tan B}{1-\tan A\tan B}=\frac{1}{\sqrt{3}}.$

Let $x=\tan A,\quad y=\tan B.$ Then $\frac{x+y}{1-xy}=\frac{1}{\sqrt{3}}.$

Express $x+y$ in terms of $xy$

Rearranging, $x+y=\frac{1-xy}{\sqrt{3}}.$ But this is not directly enough for minimization.

Use symmetry / Jensen / convexity idea

Since $A,B>0$ and $A+B$ is fixed, consider $f(t)=\tan t.$ On $\left(0,\frac{\pi}{6}\right)$, $f''(t)=2\sec^2 t\tan t>0,$ so $\tan t$ is convex.

For a convex function, with fixed sum $A+B$, the expression $\tan A+\tan B$ is minimized when $A=B=\frac{A+B}{2}=\frac{\pi}{12}.$

Compute the minimum value

$\tan A+\tan B \ge 2\tan\left(\frac{\pi}{12}\right).$

Now, $\tan\left(\frac{\pi}{12}\right)=\tan 15^\circ=2-\sqrt{3}.$

Hence, $\min(\tan A+\tan B)=2(2-\sqrt{3})=4-2\sqrt{3}.$

Check options

$4-2\sqrt{3}$ corresponds to Option C.

Comparison with stored answer

Stored correct answer: C

Derived answer: C

So they agree.