(i) Start with \(\left( \frac{1}{\sin \theta} - \frac{1}{\tan \theta} \right)^2\).

Rewrite \(\frac{1}{\tan \theta}\) as \(\frac{\cos \theta}{\sin \theta}\), so:

\(\left( \frac{1}{\sin \theta} - \frac{\cos \theta}{\sin \theta} \right)^2 = \left( \frac{1 - \cos \theta}{\sin \theta} \right)^2\).

This becomes \(\frac{(1 - \cos \theta)^2}{\sin^2 \theta}\).

Using \(\sin^2 \theta = 1 - \cos^2 \theta\), we have:

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

(ii) Use the result from part (i):

\(\frac{1 - \cos \theta}{1 + \cos \theta} = \frac{2}{5}\).

Cross-multiply to get:

\(5(1 - \cos \theta) = 2(1 + \cos \theta)\).

Simplify to find \(\cos \theta\):

\(5 - 5\cos \theta = 2 + 2\cos \theta\).

\(3 = 7\cos \theta\).

\(\cos \theta = \frac{3}{7}\).

Find \(\theta\) using inverse cosine:

\(\theta = 64.6^\circ\) or \(295.4^\circ\).

(i) Start with the left-hand side:

\(\frac{\cos \theta}{\tan \theta (1 - \sin \theta)} = \frac{\cos^2 \theta}{\sin \theta (1 - \sin \theta)}\)

Using \(\cos^2 \theta = 1 - \sin^2 \theta\), we have:

\(= \frac{1 - \sin^2 \theta}{\sin \theta (1 - \sin \theta)}\)

Factor the numerator:

\(= \frac{(1 - \sin \theta)(1 + \sin \theta)}{\sin \theta (1 - \sin \theta)}\)

Cancel \((1 - \sin \theta)\):

\(= \frac{1 + \sin \theta}{\sin \theta} = 1 + \frac{1}{\sin \theta}\)

Thus, the identity is proven.

(ii) Solve \(\frac{\cos \theta}{\tan \theta (1 - \sin \theta)} = 4\):

From part (i), \(1 + \frac{1}{\sin \theta} = 4\).

\(\frac{1}{\sin \theta} = 3\)

\(\sin \theta = \frac{1}{3}\)

Find \(\theta\) in the range \(0^\circ \leq \theta \leq 360^\circ\):

\(\theta = \arcsin \left( \frac{1}{3} \right) \approx 19.5^\circ\)

Also, \(\theta = 180^\circ - 19.5^\circ = 160.5^\circ\).

Thus, \(\theta = 19.5^\circ, 160.5^\circ\).

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

(i) Start with the equation:

\(3(2 \sin x - \cos x) = 2(\sin x - 3 \cos x)\)

Expand both sides:

\(6 \sin x - 3 \cos x = 2 \sin x - 6 \cos x\)

Rearrange to collect like terms:

\(6 \sin x - 2 \sin x = -6 \cos x + 3 \cos x\)

\(4 \sin x = -3 \cos x\)

Divide both sides by \(\cos x\):

\(\frac{4 \sin x}{\cos x} = -3\)

\(\tan x = -\frac{3}{4}\)

(ii) From \(\tan x = -\frac{3}{4}\), find the general solution:

\(x = \tan^{-1}\left(-\frac{3}{4}\right)\)

\(x \approx -36.9^\circ\)

Adjust for the range \(0^\circ \leq x \leq 360^\circ\):

\(x = 180^\circ - 36.9^\circ = 143.1^\circ\)

\(x = 360^\circ - 36.9^\circ = 323.1^\circ\)

(i) Expand the left-hand side: \((\sin x + \cos x)(1 - \sin x \cos x) = \sin x + \cos x - \sin^2 x \cos x - \cos^2 x \sin x\).

Rearrange terms: \(= \sin x + \cos x - \sin x \cos^2 x - \cos x \sin^2 x\).

Use identities: \(\sin^2 x = 1 - \cos^2 x\) and \(\cos^2 x = 1 - \sin^2 x\).

Substitute: \(\sin x \cos^2 x = \sin x (1 - \sin^2 x) = \sin x - \sin^3 x\) and \(\cos x \sin^2 x = \cos x (1 - \cos^2 x) = \cos x - \cos^3 x\).

Combine: \(\sin x + \cos x - (\sin x - \sin^3 x) - (\cos x - \cos^3 x) = \sin^3 x + \cos^3 x\).

Thus, the identity is proven.

(ii) Given equation: \((\sin x + \cos x)(1 - \sin x \cos x) = 9 \sin^3 x\).

Use part (i): \(\sin^3 x + \cos^3 x = 9 \sin^3 x\).

Rearrange: \(\cos^3 x = 8 \sin^3 x\).

Divide by \(\cos^3 x\): \(\tan^3 x = \frac{1}{8}\).

Take cube root: \(\tan x = \frac{1}{2}\).

Find solutions: \(x = \tan^{-1}(\frac{1}{2}) \approx 26.6^\circ\).

Use periodicity: \(x = 26.6^\circ + 180^\circ = 206.6^\circ\).

Thus, the solutions are \(x = 26.6^\circ\) and \(x = 206.6^\circ\).