(i) Start with \(\left( \frac{1}{\sin x} - \frac{1}{\tan x} \right)^2 = \left( \frac{s - c}{s} \right)^2\) where \(s = \sin x\) and \(c = \cos x\).

\(= \frac{(1-c)^2}{s^2} = \frac{(1-c)^2}{1-c^2}\) using \(s^2 = 1-c^2\).

\(= \frac{(1-c)(1-c)}{(1-c)(1+c)} = \frac{1-c}{1+c}\).

Thus, \(\left( \frac{1}{\sin x} - \frac{1}{\tan x} \right)^2 \equiv \frac{1 - \cos x}{1 + \cos x}\).

(ii) Given \(\left( \frac{1}{\sin x} - \frac{1}{\tan x} \right)^2 = \frac{2}{5}\), equate to \(\frac{1-c}{1+c} = \frac{2}{5}\).

Solving for \(c\), \(\frac{1-c}{1+c} = \frac{2}{5}\) gives \(5(1-c) = 2(1+c)\).

\(5 - 5c = 2 + 2c\) leads to \(3 = 7c\) or \(c = \frac{3}{7}\).

Thus, \(\cos x = \frac{3}{7}\).

Solving for \(x\), \(x = \cos^{-1} \left( \frac{3}{7} \right) \approx 1.13\) or \(x = 2\pi - 1.13 \approx 5.16\).