Answer: The stationary point is \((\frac{\pi}{6},\sqrt{3})\), and it is a minimum point.

Rewrite the function as

\(y=\frac{2-\sin x}{\cos x}=2\sec x-\tan x.\)

Differentiate:

\(\frac{dy}{dx}=2\sec x\tan x-\sec^2x.\)

For a stationary point, set \(\dfrac{dy}{dx}=0\):

\(2\sec x\tan x-\sec^2x=0.\)

Factorising,

\(\sec x(2\tan x-\sec x)=0.\)

Since \(\sec x\ne0\),

\(2\tan x=\sec x.\)

Using \(\tan x=\dfrac{\sin x}{\cos x}\) and \(\sec x=\dfrac{1}{\cos x}\),

\(2\sin x=1.\)

Hence

\(\sin x=\frac12.\)

In the interval \(-\dfrac{\pi}{2}<x<\dfrac{\pi}{2}\), this gives

\(x=\frac{\pi}{6}.\)

Substitute into the curve:

\(y=\frac{2-\frac12}{\frac{\sqrt3}{2}}=\sqrt3.\)

So the stationary point is

\(\left(\frac{\pi}{6},\sqrt3\right).\)

To determine its nature, check the sign of \(\dfrac{dy}{dx}\). For example, at \(x=0\),

\(2\sec x\tan x-\sec^2x=-1<0,\)

while for a value just greater than \(\dfrac{\pi}{6}\), the derivative is positive. Therefore the gradient changes from negative to positive, so the stationary point is a minimum.