[![Source: Wikimedia Commoms](https://i.stack.imgur.com/IbRXK.png)](https://i.stack.imgur.com/IbRXK.png)

Suppose we are given the free-body diagram above, with a mass on an incline at an angle $\theta$. If my coordinate axes are taken to be the regular $x-y$ plane rotated an angle of $\theta$ (*i.e.* the x axis is parallel to the direction of $mg \sin \theta$ and the $y$ axis is parallel to $mg \cos \theta$), then if there is no vertical motion, $N = mg \cos \theta$.

Now suppose I take my coordinate axes to be the $x-y$ plane but **NOT** rotated by any angle (*i.e.* the $x$ axis is parallel to the base of the triangle and the $y$ axis is parallel to the adjacent side), then if I want to resolve my weight $mg$ in the direction of $N$, wouldn't I need to make it $\frac{mg}{\cos \theta}$? (since in this case I would have to find the force whose vertical component is $mg$) If this is the case, no vertical motion would imply that $N = \frac{mg}{\cos \theta}$.

Does this mean that $N$ depends on my choice of coordinate axes? I might be making a mistake, but I'd appreciate if someone could point out the mistake in my logic.