If $0\le \mathrm{x}\le \frac{\mathrm{\pi }}{3}$ then range of  $\mathrm{f}\left(\mathrm{x}\right)=\sec \left(\frac{\mathrm{\pi }}{6}-\mathrm{x}\right)+\sec \left(\frac{\mathrm{\pi }}{6}+\mathrm{x}\right)$ is

if a,b>0

Using A.M. $\ge$ G.M., we get 

$\begin{array}{l}\frac{1}{\mathrm{a}}+\frac{1}{\mathrm{b}}\ge \frac{2}{\sqrt{\mathrm{ab}}}\\ \Rightarrow \mathrm{f}\left(\mathrm{x}\right)\ge \frac{2}{\sqrt{\cos \left(\frac{\mathrm{\pi }}{6}-\mathrm{x}\right)\cos \left(\frac{\mathrm{\pi }}{6}+\mathrm{x}\right)}}\\ =\frac{2}{\sqrt{{\cos }^2\frac{\mathrm{\pi }}{6}-{\sin }^2\mathrm{x}}}\\ =\frac{2}{\sqrt{\frac{3}{4}-\frac{1-\cos 2\mathrm{x}}{2}}}\\ =\frac{2}{\sqrt{\frac{1}{4}+\frac{\cos 2\mathrm{x}}{2}}}\end{array}$

$\text{ Now for }0\le \mathrm{x}\le \frac{\mathrm{\pi }}{3},\frac{-1}{2}\le \cos 2\mathrm{x}\le 1$

$\begin{array}{l}\Rightarrow 0\le \sqrt{\frac{1}{4}+\frac{\cos 2\mathrm{x}}{2}}\le \frac{\sqrt{3}}{2}\\ \Rightarrow \mathrm{f}\left(\mathrm{x}\right)\ge \frac{4}{\sqrt{3}}\end{array}$

Since 'f is continuous, range of ${\mathrm{f}}^{\mathrm{\prime }}\text{ is }\left[\frac{4}{\sqrt{3}},\mathrm{\infty }\right)\text{. }$