(i) Start with \(\frac{\tan^2 \theta - 1}{\tan^2 \theta + 1} = \frac{\frac{\sin^2 \theta}{\cos^2 \theta} - 1}{\frac{\sin^2 \theta}{\cos^2 \theta} + 1}\).

Multiply numerator and denominator by \(\cos^2 \theta\):

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

Using \(\sin^2 \theta + \cos^2 \theta = 1\), this simplifies to \(\sin^2 \theta - \cos^2 \theta = \sin^2 \theta - (1 - \sin^2 \theta) = 2\sin^2 \theta - 1\).

Thus, \(a = 2\) and \(b = -1\).

(ii) Solve \(2\sin^2 \theta - 1 = \frac{1}{4}\).

\(2\sin^2 \theta = \frac{1}{4} + 1 = \frac{5}{4}\).

\(\sin^2 \theta = \frac{5}{8}\).

\(\sin \theta = \pm \sqrt{\frac{5}{8}} \approx \pm 0.7906\).

Since \(-90^\circ \leq \theta \leq 0^\circ\), \(\theta = -52.2^\circ\).

(i) To prove the identity \((\sin \theta + \cos \theta)(1 - \sin \theta \cos \theta) \equiv \sin^3 \theta + \cos^3 \theta\):

Expand the left-hand side (LHS):

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

Rearrange terms:

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

Using \(\sin^2 \theta + \cos^2 \theta = 1\), this simplifies to:

\(= \sin^3 \theta + \cos^3 \theta\)

Thus, the identity is proven.

(ii) Solve \((\sin \theta + \cos \theta)(1 - \sin \theta \cos \theta) = 3 \cos^3 \theta\):

From part (i), \(\sin^3 \theta + \cos^3 \theta = 3 \cos^3 \theta\)

Rearrange to get:

\(\sin^3 \theta = 2 \cos^3 \theta\)

Divide both sides by \(\cos^3 \theta\):

\(\tan^3 \theta = 2\)

Taking the cube root gives:

\(\tan \theta = \sqrt[3]{2}\)

Calculate \(\theta\) using \(\tan^{-1}(\sqrt[3]{2})\):

\(\theta = 51.6^\circ\) or \(231.6^\circ\) (considering the periodicity of tangent).

(i) Start with the given equation:

\(\frac{2 \sin \theta + \cos \theta}{\sin \theta + \cos \theta} = 2 \tan \theta\)

Replace \(\tan \theta\) with \(\frac{\sin \theta}{\cos \theta}\):

\(\frac{2 \sin \theta + \cos \theta}{\sin \theta + \cos \theta} = 2 \cdot \frac{\sin \theta}{\cos \theta}\)

Multiply both sides by \(\cos \theta (\sin \theta + \cos \theta)\):

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

Expand both sides:

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

Subtract \(2 \sin \theta \cos \theta\) from both sides:

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

(ii) From \(\cos^2 \theta = 2 \sin^2 \theta\), use the identity \(\sin^2 \theta + \cos^2 \theta = 1\):

\(\cos^2 \theta = 2 (1 - \cos^2 \theta)\)

\(\cos^2 \theta = 2 - 2 \cos^2 \theta\)

\(3 \cos^2 \theta = 2\)

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

\(\cos \theta = \pm \sqrt{\frac{2}{3}}\)

For \(0^\circ < \theta < 180^\circ\), \(\theta = 35.3^\circ\) or \(144.7^\circ\).

(i) Start with the left-hand side: \(\left( \frac{1}{\cos \theta} - \tan \theta \right)^2\).

Rewrite \(\tan \theta\) as \(\frac{\sin \theta}{\cos \theta}\), so the expression becomes \(\left( \frac{1 - \sin \theta}{\cos \theta} \right)^2\).

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

Using the identity \(\cos^2 \theta = 1 - \sin^2 \theta\), the expression becomes \(\frac{(1 - \sin \theta)^2}{1 - \sin^2 \theta}\).

Factor the denominator: \(1 - \sin^2 \theta = (1 - \sin \theta)(1 + \sin \theta)\).

Cancel \(1 - \sin \theta\) from the numerator and denominator to get \(\frac{1 - \sin \theta}{1 + \sin \theta}\), which matches the right-hand side.

(ii) From part (i), we have \(\left( \frac{1}{\cos \theta} - \tan \theta \right)^2 = \frac{1 - \sin \theta}{1 + \sin \theta}\).

Set \(\frac{1 - \sin \theta}{1 + \sin \theta} = \frac{1}{2}\).

Cross-multiply to get \(2(1 - \sin \theta) = 1 + \sin \theta\).

Simplify to \(2 - 2\sin \theta = 1 + \sin \theta\).

Rearrange to \(2 - 1 = 2\sin \theta + \sin \theta\).

Thus, \(1 = 3\sin \theta\) or \(\sin \theta = \frac{1}{3}\).

Find \(\theta\) using \(\sin^{-1} \left( \frac{1}{3} \right)\), which gives \(\theta = 19.5^\circ\) or \(\theta = 160.5^\circ\) within the range \(0^\circ \leq \theta \leq 360^\circ\).

(i) Start with the left-hand side: \(\frac{1 + \cos \theta}{\sin \theta} + \frac{\sin \theta}{1 + \cos \theta}\).

Combine the fractions: \(\frac{(1 + \cos \theta)^2 + \sin^2 \theta}{\sin \theta (1 + \cos \theta)}\).

Use the identity \(\sin^2 \theta = 1 - \cos^2 \theta\):

\((1 + \cos \theta)^2 + \sin^2 \theta = 1 + 2\cos \theta + \cos^2 \theta + 1 - \cos^2 \theta = 2 + 2\cos \theta\).

So, \(\frac{2 + 2\cos \theta}{\sin \theta (1 + \cos \theta)} = \frac{2(1 + \cos \theta)}{\sin \theta (1 + \cos \theta)} = \frac{2}{\sin \theta}\).

This proves the identity.

(ii) Use the result from part (i): \(\frac{2}{\sin \theta} = \frac{3}{\cos \theta}\).

Cross-multiply: \(2\cos \theta = 3\sin \theta\).

Divide both sides by \(\cos \theta \sin \theta\): \(\frac{2}{\sin \theta} = \frac{3}{\cos \theta} \rightarrow t = \frac{2}{3}\).

\(\tan \theta = \frac{2}{3}\).

Find \(\theta\) using inverse tangent: \(\theta = \tan^{-1}\left(\frac{2}{3}\right) \approx 33.7^\circ\).

Since \(\tan \theta\) is positive in the first and third quadrants, the solutions are \(\theta = 33.7^\circ\) and \(\theta = 180^\circ + 33.7^\circ = 213.7^\circ\).