If $\tan \theta +\sec \theta =4$, then what is the value of $\cos \theta$?

We know the fundamental trigonometric identity:
$${\sec }^2\theta -{\tan }^2\theta =1$$

This can be factored using the difference of squares identity:

$$(\sec \theta -\tan \theta )(\sec \theta +\tan \theta )=1$$

Given that $\sec \theta +\tan \theta =4$, substituting this into the equation yields:

$$(\sec \theta -\tan \theta )\times 4=1⟹\sec \theta -\tan \theta =\frac{1}{4}$$

Now, add the two equations together:

$$(\sec \theta +\tan \theta )+(\sec \theta -\tan \theta )=4+\frac{1}{4}$$

$$2\sec \theta =\frac{17}{4}⟹\sec \theta =\frac{17}{8}$$

Since $\cos \theta =\frac{1}{\sec \theta }$:

$$\cos \theta =\frac{8}{17}$$