How to properly compute the stable vertical path load factor I've been told by someone else that in a stable flight when the vertical path of the aircraft is maintained, the Nz value is equal to cos(pitch)/cos(bank), but I can't seem to figure out how that is derived, the only place on the whole internet where I was able to find that info was here: Inside another stack exchange post, however, it didn't explain how was the formula derived. I do understand that inside a banked turn with 0 disturbance to the vertical path the aircraft's load factor is defined by 1/cos(bank) as explained in the textbook I read: Introduction to Aerospace Engineering with a Flight Test Perspective. 
But this doesn't explain the cos(pitch) part of the formula, and nowhere was I able to find anything related to it on the internet. Can someone please explain this to me? Thanks!
Let's look at the extreme case of a 90° pitch attitude. The airplane climbs vertically. Thankfully, in case of an F-16 without loads this is possible (left side of sketch below).

The vertical load factor is defined in the body axis system (blue arrows): $n_z$ is the multiple of gravitational acceleration that presses the pilot into his seat. With $\Theta$ = 90° no gravitational acceleration along the body z axis is left (red arrow), all acts along the body x axis. So here $n_z$ is zero and $n_x$ is -1. Gravitational acceleration presses the pilot into his backrest, but his behind is unloaded.
Since the cosine is 1 for 0° and 0 for 90°, it describes the variation in $n_z$ over pitch. For 180° the cosine becomes -1 and the pilot will hang in his shoulder straps. And so on. The right side of the sketch shows an intermediate case and it should be immediately clear that the component of g in z direction is sin(45°) = 1/√2.
Combine this with the change in $n_z$ for a turn and the formulas
$$n_x = -\frac{sin\Theta}{cos\phi}$$
$$n_z = \frac{cos\Theta}{cos\phi}$$
should make sense. Since the x axis is defined as positive in flight direction, the x component of the load factor is negative with positive pitch attitude, hence the sign of the equation for $n_x$.