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(B) Given: $XZ - YZ = 2$ and $XY = 2\sqrt{6}$.In right-angled triangle $XYZ$,by Pythagoras theorem:$XY^2 + YZ^2 = XZ^2$$(2\sqrt{6})^2 = XZ^2 - YZ^2$$2

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$\sqrt{6}$

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(B) Given: $XZ - YZ = 2$ and $XY = 2\sqrt{6}$.In right-angled triangle $XYZ$,by Pythagoras theorem:$XY^2 + YZ^2 = XZ^2$$(2\sqrt{6})^2 = XZ^2 - YZ^2$$24 = (XZ - YZ)(XZ + YZ)$Since $XZ - YZ = 2$,we have:$24 = 2(XZ + YZ)$$XZ + YZ = 12$Now,we have a system of two linear equations:$1$) $XZ - YZ = 2$$2$) $XZ + YZ = 12$Adding equations $(1)$ and $(2)$:$2XZ = 14 \Rightarrow XZ = 7$Substituting $XZ = 7$ in equation $(1)$:$7 - YZ = 2 \Rightarrow YZ = 5$Now,for angle $X$:$\sec X = \frac{	ext{Hypotenuse}}{	ext{Adjacent side}} = \frac{XZ}{XY} = \frac{7}{2\sqrt{6}}$$	an X = \frac{	ext{Opposite side}}{	ext{Adjacent side}} = \frac{YZ}{XY} = \frac{5}{2\sqrt{6}}$Therefore,$\sec X + 	an X = \frac{7}{2\sqrt{6}} + \frac{5}{2\sqrt{6}} = \frac{12}{2\sqrt{6}} = \frac{6}{\sqrt{6}} = \sqrt{6}$.

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

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