(i) Use the tangent addition and subtraction formulas:

\(\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}\)

\(\tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}\)

Substitute \(A = 30^\circ\), \(B = \theta\), and \(A = 60^\circ\), \(B = \theta\):

\(\tan(30^\circ + \theta) = \frac{\tan 30^\circ + \tan \theta}{1 - \tan 30^\circ \tan \theta}\)

\(\tan(60^\circ - \theta) = \frac{\tan 60^\circ - \tan \theta}{1 + \tan 60^\circ \tan \theta}\)

Given \(\tan 30^\circ = \frac{1}{\sqrt{3}}\) and \(\tan 60^\circ = \sqrt{3}\), substitute these values:

\(\frac{\frac{1}{\sqrt{3}} + \tan \theta}{1 - \frac{1}{\sqrt{3}} \tan \theta} = 2 \times \frac{\sqrt{3} - \tan \theta}{1 + \sqrt{3} \tan \theta}\)

Cross-multiply and simplify:

\((\sqrt{3} + \tan \theta)(1 + \sqrt{3} \tan \theta) = 2(\sqrt{3} - \tan \theta)(1 - \frac{1}{\sqrt{3}} \tan \theta)\)

After simplification, you obtain:

\(\tan^2 \theta + 6\sqrt{3} \tan \theta - 5 = 0\)

(ii) Solve the quadratic equation:

\(\tan^2 \theta + 6\sqrt{3} \tan \theta - 5 = 0\)

Using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1\), \(b = 6\sqrt{3}\), \(c = -5\):

\(\tan \theta = \frac{-6\sqrt{3} \pm \sqrt{(6\sqrt{3})^2 - 4 \times 1 \times (-5)}}{2 \times 1}\)

\(\tan \theta = \frac{-6\sqrt{3} \pm \sqrt{108 + 20}}{2}\)

\(\tan \theta = \frac{-6\sqrt{3} \pm \sqrt{128}}{2}\)

\(\tan \theta = \frac{-6\sqrt{3} \pm 8\sqrt{2}}{2}\)

Calculate \(\theta\) for \(0^\circ \leq \theta \leq 180^\circ\):

\(\theta = 24.7^\circ\) and \(\theta = 95.3^\circ\).

(i) Use the tangent addition formula: \(\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}\).

For \(\tan(45^\circ + x)\):

\(\tan(45^\circ + x) = \frac{\tan 45^\circ + \tan x}{1 - \tan 45^\circ \tan x} = \frac{1 + \tan x}{1 - \tan x}\)

Substitute into the equation:

\(\frac{1 + \tan x}{1 - \tan x} - \tan x = 2\)

Multiply through by \(1 - \tan x\):

\(1 + \tan x - \tan x(1 - \tan x) = 2(1 - \tan x)\)

\(1 + \tan x - \tan x + \tan^2 x = 2 - 2\tan x\)

\(1 + \tan^2 x = 2 - 2\tan x\)

Rearrange to obtain:

\(\tan^2 x + 2\tan x - 1 = 0\)

(ii) Solve the quadratic equation \(\tan^2 x + 2\tan x - 1 = 0\).

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

\(\tan x = \frac{-2 \pm \sqrt{2^2 - 4 \times 1 \times (-1)}}{2 \times 1}\)

\(\tan x = \frac{-2 \pm \sqrt{4 + 4}}{2}\)

\(\tan x = \frac{-2 \pm \sqrt{8}}{2}\)

\(\tan x = \frac{-2 \pm 2\sqrt{2}}{2}\)

\(\tan x = -1 \pm \sqrt{2}\)

For \(\tan x = -1 + \sqrt{2}\), \(x = 22.5^\circ\).

For \(\tan x = -1 - \sqrt{2}\), \(x = 112.5^\circ\).

Thus, the solutions are \(x = 22.5^\circ\) and \(x = 112.5^\circ\).

(i) Use the tangent addition and subtraction formulas:

\(\tan(45^\circ + x) = \frac{1 + \tan x}{1 - \tan x}\)

\(\tan(45^\circ - x) = \frac{1 - \tan x}{1 + \tan x}\)

Substitute into the equation:

\(\frac{1 + \tan x}{1 - \tan x} = 2 \cdot \frac{1 - \tan x}{1 + \tan x}\)

Cross-multiply to clear the fractions:

\((1 + \tan x)^2 = 2(1 - \tan x)(1 + \tan x)\)

Expand both sides:

\(1 + 2\tan x + \tan^2 x = 2(1 - \tan^2 x)\)

\(1 + 2\tan x + \tan^2 x = 2 - 2\tan^2 x\)

Rearrange to form a quadratic equation:

\(\tan^2 x + 2\tan x + 1 = 2 - 2\tan^2 x\)

\(3\tan^2 x + 2\tan x - 1 = 0\)

Divide by 3:

\(\tan^2 x - 6\tan x + 1 = 0\)

(ii) Solve the quadratic equation \(\tan^2 x - 6\tan x + 1 = 0\):

Let \(y = \tan x\), then:

\(y^2 - 6y + 1 = 0\)

Use the quadratic formula:

\(y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)

\(y = \frac{6 \pm \sqrt{36 - 4}}{2}\)

\(y = \frac{6 \pm \sqrt{32}}{2}\)

\(y = \frac{6 \pm 4\sqrt{2}}{2}\)

\(y = 3 \pm 2\sqrt{2}\)

Calculate \(x\) for \(y = \tan x\):

\(x = \tan^{-1}(3 + 2\sqrt{2}) \approx 80.3^\circ\)

\(x = \tan^{-1}(3 - 2\sqrt{2}) \approx 9.7^\circ\)

(i) Start with the equation:

\(\sin(x - 60^\circ) - \cos(30^\circ - x) = 1\)

Use the identity \(\sin(a - b) = \sin a \cos b - \cos a \sin b\):

\(\sin(x - 60^\circ) = \sin x \cos 60^\circ - \cos x \sin 60^\circ\)

\(\cos(30^\circ - x) = \cos 30^\circ \cos x + \sin 30^\circ \sin x\)

Substitute these into the equation:

\((\sin x \cos 60^\circ - \cos x \sin 60^\circ) - (\cos 30^\circ \cos x + \sin 30^\circ \sin x) = 1\)

Using \(\cos 60^\circ = \frac{1}{2}\), \(\sin 60^\circ = \frac{\sqrt{3}}{2}\), \(\cos 30^\circ = \frac{\sqrt{3}}{2}\), \(\sin 30^\circ = \frac{1}{2}\):

\(\left(\sin x \cdot \frac{1}{2} - \cos x \cdot \frac{\sqrt{3}}{2}\right) - \left(\frac{\sqrt{3}}{2} \cos x + \frac{1}{2} \sin x\right) = 1\)

Simplify:

\(\sin x \cdot \frac{1}{2} - \frac{\sqrt{3}}{2} \cos x - \frac{\sqrt{3}}{2} \cos x - \frac{1}{2} \sin x = 1\)

\(-\sqrt{3} \cos x = 1\)

\(\cos x = -\frac{1}{\sqrt{3}}\)

Thus, \(k = -\frac{1}{\sqrt{3}}\).

(ii) Solve \(\cos x = -\frac{1}{\sqrt{3}}\) for \(0^\circ < x < 180^\circ\):

The angle \(x\) where \(\cos x = -\frac{1}{\sqrt{3}}\) is approximately \(x = 125.3^\circ\).

Given:

\(\tan(\alpha + \beta) = 2\)

\(\tan \alpha = 3 \tan \beta\)

Using the tangent addition formula:

\(\tan(\alpha + \beta) = \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta} = 2\)

Substitute \(\tan \alpha = 3 \tan \beta\):

\(\frac{3 \tan \beta + \tan \beta}{1 - 3 \tan^2 \beta} = 2\)

Simplify:

\(\frac{4 \tan \beta}{1 - 3 \tan^2 \beta} = 2\)

Cross-multiply:

\(4 \tan \beta = 2(1 - 3 \tan^2 \beta)\)

\(4 \tan \beta = 2 - 6 \tan^2 \beta\)

Rearrange to form a quadratic equation:

\(6 \tan^2 \beta + 4 \tan \beta - 2 = 0\)

Divide by 2:

\(3 \tan^2 \beta + 2 \tan \beta - 1 = 0\)

Let \(x = \tan \beta\), solve:

\(3x^2 + 2x - 1 = 0\)

Using the quadratic formula:

\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)

\(x = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 3 \cdot (-1)}}{2 \cdot 3}\)

\(x = \frac{-2 \pm \sqrt{4 + 12}}{6}\)

\(x = \frac{-2 \pm \sqrt{16}}{6}\)

\(x = \frac{-2 \pm 4}{6}\)

\(x = \frac{2}{6} = \frac{1}{3} \quad \text{or} \quad x = \frac{-6}{6} = -1\)

Thus, \(\tan \beta = \frac{1}{3}\) or \(\tan \beta = -1\).

For \(\tan \beta = \frac{1}{3}\), \(\tan \alpha = 1\), so \(\alpha = 45^\circ\) and \(\beta = 18.4^\circ\).

For \(\tan \beta = -1\), \(\tan \alpha = -3\), so \(\alpha = 108.4^\circ\) and \(\beta = 135^\circ\).