(i) Start with the left-hand side: \(\frac{\sin x \tan x}{1 - \cos x} = \frac{\sin x \cdot \frac{\sin x}{\cos x}}{1 - \cos x} = \frac{\sin^2 x}{\cos x (1 - \cos x)}\).

Using the identity \(\sin^2 x = 1 - \cos^2 x\), we have:

\(\frac{1 - \cos^2 x}{\cos x (1 - \cos x)} = \frac{(1 - \cos x)(1 + \cos x)}{\cos x (1 - \cos x)}\).

Cancel \(1 - \cos x\) from numerator and denominator:

\(\frac{1 + \cos x}{\cos x} = 1 + \frac{1}{\cos x}\).

Thus, the identity is proven.

(ii) From part (i), \(\frac{1}{\cos x} + 1 + 2 = 0\) simplifies to \(\frac{1}{\cos x} = -3\).

Therefore, \(\cos x = -\frac{1}{3}\).

Solving \(\cos x = -\frac{1}{3}\) for \(0^\circ \leq x \leq 360^\circ\), we find:

\(x = 109.5^\circ\) or \(x = 250.5^\circ\).