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

In the right-angled triangle $XYZ$, $\angle Y=90^{\circ }$, meaning $XZ$ is the hypotenuse. According to Pythagoras' theorem:
$$X{Z}^2=X{Y}^2+Y{Z}^2$$

We are given $XY=2\sqrt{6}$ and $XZ-YZ=2⟹XZ=YZ+2$. Substituting these into the identity:

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

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

$$4YZ+4=24$$

$$4YZ=20⟹YZ=5$$

Since $YZ=5$, the hypotenuse $XZ$ is:

$$XZ=5+2=7$$

With respect to acute angle $X$:

* $\text{Base}=XY=2\sqrt{6}$
* $\text{Perpendicular}=YZ=5$
* $\text{Hypotenuse}=XZ=7$

Now, evaluating the expression $\sec X+\tan X$:

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

$$\tan X=\frac{\text{Perpendicular}}{\text{Base}}=\frac{5}{2\sqrt{6}}$$

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