If $A+B=\frac{\mathrm{\pi }}{3}$where $A,B\in R^{+}$.  Then the minimum value of $\sec A+\sec B$ is

For $y=\sec x,x\in \left(0,\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)$

$Tangent drawn to it at any point lies completely below the graph of$$y=\sec x$

$\therefore \frac{\sec A+\sec B}{2}\ge \sec \left(\frac{A+B}{2}\right)$

$\Rightarrow \sec A+\sec B\ge 2.\sec \left(\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$6$}\right.\right)=\frac{4}{\sqrt{3}}$