For the curve shown in the diagram, the normal to the curve at the point \(P\) with coordinates \((x, y)\) meets the \(x\)-axis at \(N\). The point \(M\) is the foot of the perpendicular from \(P\) to the \(x\)-axis.

The curve is such that for all values of \(x\) in the interval \(0 \leq x < \frac{1}{2}\pi\), the area of triangle \(PMN\) is equal to \(\tan x\).

(a) (i) Show that \(\frac{MN}{y} = \frac{dy}{dx}\).

(ii) Hence show that \(x\) and \(y\) satisfy the differential equation \(\frac{1}{2}y^2 \frac{dy}{dx} = \tan x\).

(b) Given that \(y = 1\) when \(x = 0\), solve this differential equation to find the equation of the curve, expressing \(y\) in terms of \(x\).