(i) Start with the equation \(3 \tan \theta = 2 \cos \theta\).

Using \(\tan \theta = \frac{\sin \theta}{\cos \theta}\), rewrite as \(3 \frac{\sin \theta}{\cos \theta} = 2 \cos \theta\).

Multiply through by \(\cos \theta\) to get \(3 \sin \theta = 2 \cos^2 \theta\).

Use the identity \(\cos^2 \theta = 1 - \sin^2 \theta\) to substitute: \(3 \sin \theta = 2(1 - \sin^2 \theta)\).

Expand and rearrange: \(3 \sin \theta = 2 - 2 \sin^2 \theta\).

Rearrange to form the quadratic: \(2 \sin^2 \theta + 3 \sin \theta - 2 = 0\).

(ii) Solve the quadratic equation \(2 \sin^2 \theta + 3 \sin \theta - 2 = 0\).

Let \(s = \sin \theta\). The equation becomes \(2s^2 + 3s - 2 = 0\).

Factorize to find \(s = 0.5\) or \(s = -2\).

Since \(-1 \leq \sin \theta \leq 1\), only \(s = 0.5\) is valid.

Thus, \(\sin \theta = 0.5\) gives \(\theta = 30^\circ\) or \(150^\circ\) within the range \(0^\circ \leq \theta \leq 360^\circ\).

(i) We start with \(\sin x \tan x = \sin x \cdot \frac{\sin x}{\cos x}\).

This simplifies to \(\frac{\sin^2 x}{\cos x}\).

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

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

Thus, \(\sin x \tan x = \frac{1 - \cos^2 x}{\cos x}\).

(ii) Given \(2 \sin x \tan x = 3\), substitute \(\sin x \tan x = \frac{1 - \cos^2 x}{\cos x}\):

\(2 \cdot \frac{1 - \cos^2 x}{\cos x} = 3\).

Multiply through by \(\cos x\):

\(2(1 - \cos^2 x) = 3 \cos x\).

Expand and rearrange:

\(2 - 2\cos^2 x = 3\cos x\).

\(2\cos^2 x + 3\cos x - 2 = 0\).

Let \(c = \cos x\), then:

\(2c^2 + 3c - 2 = 0\).

Using the quadratic formula \(c = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 2, b = 3, c = -2\):

\(c = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 2 \cdot (-2)}}{2 \cdot 2}\).

\(c = \frac{-3 \pm \sqrt{9 + 16}}{4}\).

\(c = \frac{-3 \pm 5}{4}\).

\(c = \frac{2}{4} = 0.5\) or \(c = \frac{-8}{4} = -2\).

Since \(\cos x\) must be between -1 and 1, \(\cos x = 0.5\).

Thus, \(x = 60^\circ\) or \(x = 300^\circ\).

Start with the equation \(\tan \theta \sin \theta = 1\).

Using \(\tan \theta = \frac{\sin \theta}{\cos \theta}\), we have:

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

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

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

Using the identity \(\sin^2 \theta = 1 - \cos^2 \theta\), substitute:

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

Rearrange to form a quadratic equation:

\(\cos^2 \theta + \cos \theta - 1 = 0\).

Use the quadratic formula \(\cos \theta = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) where \(a = 1, b = 1, c = -1\):

\(\cos \theta = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot (-1)}}{2 \cdot 1}\).

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

Calculate the possible values for \(\cos \theta\):

\(\cos \theta = \frac{-1 + \sqrt{5}}{2} \approx 0.618\).

Find \(\theta\) using \(\cos^{-1}(0.618)\):

\(\theta \approx 51.8^\circ\) and \(\theta \approx 360^\circ - 51.8^\circ = 308.2^\circ\).

Start with the equation \(8 \sin^2 \theta + 6 \cos \theta + 1 = 0\).

Using the identity \(\sin^2 \theta = 1 - \cos^2 \theta\), substitute to get:

\(8(1 - \cos^2 \theta) + 6 \cos \theta + 1 = 0\).

Simplify to:

\(8 - 8 \cos^2 \theta + 6 \cos \theta + 1 = 0\).

Combine like terms:

\(-8 \cos^2 \theta + 6 \cos \theta + 9 = 0\).

Rearrange to:

\(8 \cos^2 \theta - 6 \cos \theta - 9 = 0\).

Factor the quadratic equation:

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

Set each factor to zero:

\(4 \cos \theta + 3 = 0\) or \(2 \cos \theta - 3 = 0\).

Solve for \(\cos \theta\):

\(\cos \theta = -\frac{3}{4}\).

Find \(\theta\) using the inverse cosine function:

\(\theta = \cos^{-1}(-0.75)\).

Within the range \(0^\circ < \theta < 180^\circ\), \(\theta = 138.6^\circ\).

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

Combine the fractions: \(\frac{\sin \theta (\sin \theta - \cos \theta) + \cos \theta (\sin \theta + \cos \theta)}{(\sin \theta + \cos \theta)(\sin \theta - \cos \theta)}\).

Simplify the numerator: \(\sin^2 \theta - \sin \theta \cos \theta + \cos \theta \sin \theta + \cos^2 \theta = \sin^2 \theta + \cos^2 \theta = 1\).

The denominator becomes \(\sin^2 \theta - \cos^2 \theta\).

Thus, the expression simplifies to \(\frac{1}{\sin^2 \theta - \cos^2 \theta}\).

Using the identity \(\tan^2 \theta = \frac{\sin^2 \theta}{\cos^2 \theta}\), rewrite as \(\frac{1}{\frac{\sin^2 \theta - \cos^2 \theta}{\cos^2 \theta}} = \frac{\cos^2 \theta}{\sin^2 \theta - \cos^2 \theta}\).

Recognize \(\sin^2 \theta - \cos^2 \theta = (\tan^2 \theta + 1) - 2\), so \(\frac{\cos^2 \theta}{\sin^2 \theta - \cos^2 \theta} = \frac{\tan^2 \theta + 1}{\tan^2 \theta - 1}\).

Thus, the identity is proven.

(b) Set \(\frac{\tan^2 \theta + 1}{\tan^2 \theta - 1} = 2\).

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

Simplify: \(\tan^2 \theta + 1 = 2\tan^2 \theta - 2\).

Rearrange: \(\tan^2 \theta = 3\).

Thus, \(\tan \theta = \pm \sqrt{3}\).

For \(\tan \theta = \sqrt{3}\), \(\theta = \frac{\pi}{3}\).

For \(\tan \theta = -\sqrt{3}\), \(\theta = \frac{2\pi}{3}\).

Therefore, the solutions are \(\theta = \frac{1}{3} \pi, \frac{2}{3} \pi\).