newtonian mechanics - Physical reason of difference between normal force in a banked curve and an inclined plane.

When determining the centripetal force on an object on a banked curve, it is stated that the banking angle for a given speed and radius is found by :

tanθ = v^2/rg

It is found as follows: The normal force on the object is resolved into components. The x-component (the one providing the centripetal force) is:

N*sinθ = mv^2/r

Then, the y component is set equal to mg:

N*cosθ = mg =>N = mg/cosθ

Dividing Nsinθ by Ncosθ, we get :

tanθ = v^2/rg

I'm fine with that.

Here is where I'm confused. When resolving the forces of an object resting on an inclined plane: The component down and parallel to the plane due to gravity is:

mg*sinθ.

The component representing the force of gravity into (perpendicular) to the plane is:

mg*cosθ.

The normal force is equal to this component into the plane by Newton's 3rd Law, so,

N =mg*cosθ.

Why in the first scenario (banked curve) is

N=mg/cosθ,

HOWEVER, in the second (inclined plane)

N=mg*cosθ ?

How can this be? There are two different values for N?

Is the normal force in the first scenario (banked curve) greater than mgcosθ because:

a) the normal force also is responsible for the centripetal acceleration, so it needs to be greater?

Or,

b) the car is not sliding down the curve, so the normal force is greater because of the translational equilibrium requirement? The textbook that I have (Resnick, Fundamentals of Physics) seems to suggest (b), since they use the equation for equilibrium in the Y direction. But if that is the case, then what is PHYSICALLY CREATING this "extra" normal force as compared to the second scenario (inclined plane)?