$\Delta DEF$ is right-angled at $E$. If $\cos D=\frac{8}{17}$, then what is the value of $\sec F$?

Since $\Delta DEF$ is right-angled at $E$, the angles $D$ and $F$ are complementary angles:
$$D+F=90^{\circ }⟹D=90^{\circ }-F$$
Using the co-function trigonometric identity:
$$\cos D=\cos (90^{\circ }-F)=\sin F$$
Given $\cos D=\frac{8}{17}$, it implies that $\sin F=\frac{8}{17}$.In a right-angled triangle with respect to angle $F$:
$$\sin F=\frac{\text{Perpendicular}}{\text{Hypotenuse}}=\frac{8}{17}$$
We need to find $\sec F$:
$$\sec F=\frac{\text{Hypotenuse}}{\text{Base}}$$
Using Pythagoras theorem:
$$\text{Base}=\sqrt{{\text{Hypotenuse}}^2-{\text{Perpendicular}}^2}=\sqrt{17^2-8^2}=\sqrt{289-64}=\sqrt{225}=15$$
Therefore:
$$\sec F=\frac{17}{15}$$