If A > 0, B > 0 and A + B = π/6, then the minimum value of tanA + tanB is:

We are given that A > 0, B > 0, and A + B = π/6. We want to find the minimum value of tanA + tanB.

We know that tan(A + B) = (tanA + tanB) / (1 - tanA tanB).

Since A + B = π/6, we have tan(A + B) = tan(π/6) = 1/√3.

Therefore, 1/√3 = (tanA + tanB) / (1 - tanA tanB).

Let x = tanA + tanB. Then x(1 - tanA tanB) = 1/√3.

We have (tanA + tanB)² = tan²A + 2tanA tanB + tan²B.

By AM-GM inequality, tanA + tanB ≥ 2√(tanA tanB).

Let's use the identity tan(A+B) = (tanA + tanB)/(1 - tanA tanB).

We have tan(π/6) = 1/√3 = (tanA + tanB) / (1 - tanA tanB).

Let u = tanA and v = tanB. Then 1/√3 = (u + v) / (1 - uv).

Thus, u + v = (1 - uv)/√3.

Since A + B = π/6, we can write B = π/6 - A. Then tanB = tan(π/6 - A) = (tan(π/6) - tanA) / (1 + tan(π/6)tanA) = (1/√3 - tanA) / (1 + tanA/√3).

Let f(A) = tanA + tan(π/6 - A) = tanA + (1/√3 - tanA) / (1 + tanA/√3).

To find the minimum value, we can use calculus or consider the case when A = B = π/12.

When A = B = π/12, tanA + tanB = 2tan(π/12) = 2(2 - √3) = 4 - 2√3 ≈ 0.536.

Also, note that tan(π/12) = 2 - √3.

Therefore, the minimum value of tanA + tanB is 2(2 - √3) = 4 - 2√3.

However, this is not one of the options.

Let's use Cauchy-Schwarz inequality:
(tanA + tanB)² ≤ (1² + 1²)(tan²A + tan²B).

We have tanA + tanB ≥ 2√(tanA tanB).

Let's consider the case when A = π/6 and B = 0. Then tanA + tanB = tan(π/6) = 1/√3.

Let's consider the case when A = 0 and B = π/6. Then tanA + tanB = tan(π/6) = 1/√3.

If A = B = π/12, then tanA + tanB = 2tan(π/12) = 2(2 - √3) = 4 - 2√3 ≈ 0.535898.

The minimum value seems to be 2 - √3.