In a right-angled triangle $XYZ$, $\angle Y=90^{\circ }$. If $XY=2\sqrt{6}$ and $XZ-YZ=2$, then the value of $\sec X+\tan X$ is:

In the right-angled triangle $XYZ$, applying Pythagoras theorem gives:
$$X{Z}^2=X{Y}^2+Y{Z}^2$$

From the problem, we know $XZ-YZ=2⟹XZ=YZ+2$. Substituting $XZ$ and $XY=2\sqrt{6}$ into the theorem:

$$(YZ+2{)}^2=(2\sqrt{6}{)}^2+Y{Z}^2$$

$$Y{Z}^2+4YZ+4=24+Y{Z}^2$$

Subtracting $Y{Z}^2$ from both sides:

$$4YZ+4=24$$

$$4YZ=20⟹YZ=5$$

Now calculate the hypotenuse $XZ$:

$$XZ=5+2=7$$

With respect to $\angle X$:

* $\text{Opposite side}=YZ=5$
* $\text{Adjacent side}=XY=2\sqrt{6}$
* $\text{Hypotenuse}=XZ=7$

Find the identities $\sec X$ and $\tan X$:

$$\sec X=\frac{\text{Hypotenuse}}{\text{Adjacent}}=\frac{7}{2\sqrt{6}}$$

$$\tan X=\frac{\text{Opposite}}{\text{Adjacent}}=\frac{5}{2\sqrt{6}}$$

Summing both values together:

$$\sec X+\tan X=\frac{7}{2\sqrt{6}}+\frac{5}{2\sqrt{6}}=\frac{12}{2\sqrt{6}}=\frac{6}{\sqrt{6}}=\sqrt{6}$$