Minimize the sum of tangents when sum of angles are constants

Just check if this works:

I'm assuming $A_i > 0 \ \forall \ i$. Let $f(x) = \tan x$. Now, $f'(x) = \sec^2x, f''(x) = 2 \sec^2 x \tan x > 0$ for $\pi/2 >x>0$. Hence, $\tan x$ is convex.

Now, by Weierstrass Theorem, a convex function is maximized at its endpoints. So, the function is maximized when all but one variable is equal to $C$.

Edit: Now that the OP has been edited to find instead the minimum of the function, it is even simpler by convexity. Just employ the Jensen's Inequality to get that $$\sum_{i=1}^n \tan A_i \ge n \cdot \tan \left( \frac{A_1 + A_2 + \cdots+ A_n}{n} \right) = n \tan \frac C n$$