Since the crate moves at a constant speed, its acceleration is zero, and the net force is zero ($\sum \vec{F} = 0$). We resolve the forces into components parallel ($x$) and perpendicular ($y$) to the ramp.

The forces acting on the crate are the gravitational force ($mg$), the horizontal force ($F$), and the normal force ($N$).

Sum of forces parallel to the ramp:

$$ \sum F_x = F \cos\theta - mg \sin\theta = 0 $$

Solving for $F$ gives the magnitude of the applied horizontal force.

$$ F = mg \frac{\sin\theta}{\cos\theta} = mg \tan\theta $$

Sum of forces perpendicular to the ramp:

$$ \sum F_y = N - mg \cos\theta - F \sin\theta = 0 $$ $$ N = mg \cos\theta + F \sin\theta $$

Substitute the expression for $F$:

$$ N = mg \cos\theta + (mg \tan\theta) \sin\theta = mg \left( \cos\theta + \frac{\sin^2\theta}{\cos\theta} \right) $$

Using the common denominator $\cos\theta$ and the identity $\cos^2\theta + \sin^2\theta = 1$:

$$ N = mg \left( \frac{\cos^2\theta + \sin^2\theta}{\cos\theta} \right) = \frac{mg}{\cos\theta} $$