(i) Start with the left-hand side (LHS):

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

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

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

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

Thus, the identity is proven.

(ii) Solve \(\frac{1}{\tan \theta} = 4 \tan \theta\).

\(\tan^2 \theta = \frac{1}{4}\)

\(\tan \theta = \pm \frac{1}{2}\)

For \(\tan \theta = \frac{1}{2}\), \(\theta = 26.6^\circ\).

For \(\tan \theta = -\frac{1}{2}\), \(\theta = 153.4^\circ\).

(i) Start with the expression:

\(\frac{\sin \theta}{\sin \theta + \cos \theta} + \frac{\cos \theta}{\sin \theta - \cos \theta}\)

Combine the fractions:

\(\frac{\sin \theta (\sin \theta - \cos \theta) + \cos \theta (\sin \theta + \cos \theta)}{(\sin \theta + \cos \theta)(\sin \theta - \cos \theta)}\)

Expand the numerator:

\(\sin^2 \theta - \sin \theta \cos \theta + \cos \theta \sin \theta + \cos^2 \theta\)

Which simplifies to:

\(\sin^2 \theta + \cos^2 \theta\)

Using the identity \(\sin^2 \theta + \cos^2 \theta = 1\), the expression becomes:

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

(ii) Solve the equation:

\(\frac{1}{\sin^2 \theta - \cos^2 \theta} = 3\)

\(\sin^2 \theta - \cos^2 \theta = \frac{1}{3}\)

Using \(\sin^2 \theta + \cos^2 \theta = 1\), let \(s = \sin \theta\) and \(c = \cos \theta\):

\(s^2 - c^2 = \frac{1}{3}\)

\(s^2 + c^2 = 1\)

Subtract the equations:

\(2c^2 = \frac{2}{3}\)

\(c^2 = \frac{1}{3}\)

\(c = \pm \frac{1}{\sqrt{3}}\)

\(s^2 = \frac{2}{3}\)

\(s = \pm \sqrt{\frac{2}{3}}\)

\(\tan \theta = \pm \sqrt{2}\)

Solutions for \(\theta\) are:

\(\theta = 54.7^\circ, 125.3^\circ, 234.7^\circ, 305.3^\circ\)

Start with the equation:

\(6 \cos \theta \tan \theta - 3 \cos \theta + 4 \tan \theta - 2 = 0\)

Factorise by grouping:

\(3 \cos \theta (2 \tan \theta - 1) + 2 (2 \tan \theta - 1) = 0\)

Factor out \((2 \tan \theta - 1)\):

\((2 \tan \theta - 1)(3 \cos \theta + 2) = 0\)

Set each factor to zero:

\(2 \tan \theta - 1 = 0\) or \(3 \cos \theta + 2 = 0\)

Solve for \(\theta\):

\(\tan \theta = \frac{1}{2}\)

\(\theta = \tan^{-1}\left(\frac{1}{2}\right) \approx 26.6^\circ\)

\(\cos \theta = -\frac{2}{3}\)

\(\theta = \cos^{-1}\left(-\frac{2}{3}\right) \approx 131.8^\circ\)

Thus, the solutions are \(\theta = 26.6^\circ\) and \(\theta = 131.8^\circ\).

(i) Start with the equation \(\sin 2x + 3 \cos 2x = 0\).

Rearrange to \(\tan 2x = -3\).

Find \(2x\) using the inverse tangent: \(2x = 180^\circ - 71.6^\circ\) or \(360^\circ - 71.6^\circ\).

This gives \(2x = 108.4^\circ\) or \(2x = 288.4^\circ\).

Divide by 2 to find \(x\): \(x = 54.2^\circ\) or \(x = 144.2^\circ\).

Also consider the periodicity of tangent: \(x = 234.2^\circ\) and \(x = 324.2^\circ\).

(ii) The range \(0^\circ \leq x \leq 1080^\circ\) is three times the range \(0^\circ \leq x \leq 360^\circ\), so there are 3 times the number of solutions: 12 answers.

Given the equation \(\sin 2x = 2 \cos 2x\).

Divide both sides by \(\cos 2x\) (assuming \(\cos 2x \neq 0\)) to get:

\(\tan 2x = 2\).

Find the general solution for \(2x\):

\(2x = \tan^{-1}(2) + k \cdot 180^\circ\), where \(k\) is an integer.

Calculate \(\tan^{-1}(2) \approx 63.4^\circ\).

Thus, \(2x = 63.4^\circ\) or \(2x = 243.4^\circ\) (adding \(180^\circ\)).

Divide by 2 to find \(x\):

\(x = 31.7^\circ\) or \(x = 121.7^\circ\).

Both solutions are within the given range \(0^\circ \leq x \leq 180^\circ\).