(i) Use the tangent addition and subtraction formulas:

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

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

Given \(\tan 60^\circ = \sqrt{3}\), substitute:

\(\tan(60^\circ + \theta) = \frac{\sqrt{3} + \tan \theta}{1 - \sqrt{3} \tan \theta}\)

\(\tan(60^\circ - \theta) = \frac{\sqrt{3} - \tan \theta}{1 + \sqrt{3} \tan \theta}\)

Add the two equations:

\(\tan(60^\circ + \theta) + \tan(60^\circ - \theta) = \frac{(\sqrt{3} + \tan \theta)(1 + \sqrt{3} \tan \theta) + (\sqrt{3} - \tan \theta)(1 - \sqrt{3} \tan \theta)}{(1 - \sqrt{3} \tan \theta)(1 + \sqrt{3} \tan \theta)}\)

Simplify the numerator:

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

Simplify the denominator:

\(= 1 - 3\tan^2 \theta\)

Thus, the equation becomes:

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

Therefore, \((2\sqrt{3})(1 + \tan^2 \theta) = k(1 - 3\tan^2 \theta)\).

(ii) Set \(k = 3\sqrt{3}\) and solve:

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

Divide by \(\sqrt{3}\):

\(2(1 + \tan^2 \theta) = 3(1 - 3\tan^2 \theta)\)

Expand and simplify:

\(2 + 2\tan^2 \theta = 3 - 9\tan^2 \theta\)

\(11\tan^2 \theta = 1\)

\(\tan^2 \theta = \frac{1}{11}\)

\(\tan \theta = \pm \frac{1}{\sqrt{11}}\)

Find \(\theta\) in the interval \(0^\circ \leq \theta \leq 180^\circ\):

\(\theta = 16.8^\circ, 163.2^\circ\)

To solve \(\cos(\theta + 60^\circ) = 2 \sin \theta\), we start by using the angle addition formula for cosine:

\(\cos(\theta + 60^\circ) = \cos \theta \cos 60^\circ - \sin \theta \sin 60^\circ.\)

Substituting the values \(\cos 60^\circ = \frac{1}{2}\) and \(\sin 60^\circ = \frac{\sqrt{3}}{2}\), we get:

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

Rearranging gives:

\(\frac{1}{2} \cos \theta = 2 \sin \theta + \frac{\sqrt{3}}{2} \sin \theta.\)

Combine the terms on the right:

\(\frac{1}{2} \cos \theta = \left(2 + \frac{\sqrt{3}}{2}\right) \sin \theta.\)

Divide both sides by \(\sin \theta\) (assuming \(\sin \theta \neq 0\)):

\(\frac{1}{2} \cot \theta = 2 + \frac{\sqrt{3}}{2}.\)

Solving for \(\cot \theta\), we find:

\(\cot \theta = 4 + \sqrt{3}.\)

Thus, \(\tan \theta = \frac{1}{4 + \sqrt{3}}.\)

Calculating \(\theta\) gives:

\(\theta = \tan^{-1}\left(\frac{1}{4 + \sqrt{3}}\right).\)

This results in \(\theta \approx 9.9^\circ\) and \(\theta \approx 189.9^\circ\) within the interval \(0^\circ \leq \theta \leq 360^\circ\).

To solve \(\tan(45^\circ - x) = 2 \tan x\), use the tangent subtraction formula:

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

Set this equal to \(2 \tan x\):

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

Cross-multiply to obtain:

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

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

Rearrange to form a quadratic equation:

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

Let \(y = \tan x\). Solve the quadratic equation:

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

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

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

\(y = \frac{-3 \pm \sqrt{9 + 8}}{4}\)

\(y = \frac{-3 \pm \sqrt{17}}{4}\)

Calculate the values:

\(y_1 = \frac{-3 + \sqrt{17}}{4} \approx 0.281\)

\(y_2 = \frac{-3 - \sqrt{17}}{4} \approx -1.781\)

Find \(x\) using \(\tan x = y\):

\(x_1 = \tan^{-1}(0.281) \approx 15.7^\circ\)

\(x_2 = \tan^{-1}(-1.781) \approx 119.3^\circ\)

Thus, the solutions are \(x = 15.7^\circ\) and \(x = 119.3^\circ\).

To solve the problem, we follow these steps:

Part (i): Find \(\sin(a - 30^\circ)\)

We know \(\cos a = \frac{3}{5}\). Using the Pythagorean identity, \(\sin a = \sqrt{1 - \cos^2 a} = \sqrt{1 - \left(\frac{3}{5}\right)^2} = \frac{4}{5}\).

Using the angle subtraction formula for sine:

\(\sin(a - 30^\circ) = \sin a \cos 30^\circ - \cos a \sin 30^\circ\)

Substitute the known values:

\(\sin(a - 30^\circ) = \frac{4}{5} \cdot \frac{\sqrt{3}}{2} - \frac{3}{5} \cdot \frac{1}{2}\)

\(= \frac{4\sqrt{3}}{10} - \frac{3}{10}\)

\(= \frac{1}{10}(4\sqrt{3} - 3)\)

Part (ii): Find \(\tan 2a\) and \(\tan 3a\)

Using the double angle formula for tangent:

\(\tan 2a = \frac{2 \tan a}{1 - \tan^2 a}\)

First, find \(\tan a = \frac{\sin a}{\cos a} = \frac{4/5}{3/5} = \frac{4}{3}\).

Substitute into the formula:

\(\tan 2a = \frac{2 \cdot \frac{4}{3}}{1 - \left(\frac{4}{3}\right)^2} = \frac{\frac{8}{3}}{1 - \frac{16}{9}}\)

\(= \frac{\frac{8}{3}}{-\frac{7}{9}} = -\frac{24}{7}\)

Using the angle addition formula for tangent:

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

Substitute \(\tan 2a = -\frac{24}{7}\) and \(\tan a = \frac{4}{3}\):

\(\tan 3a = \frac{-\frac{24}{7} + \frac{4}{3}}{1 - \left(-\frac{24}{7}\right) \left(\frac{4}{3}\right)}\)

\(= \frac{-\frac{72}{21} + \frac{28}{21}}{1 + \frac{96}{21}}\)

\(= \frac{-\frac{44}{21}}{\frac{117}{21}} = -\frac{44}{117}\)

1. Given \(\tan \alpha = 2 \tan \beta\) and \(\tan(\alpha + \beta) = 3\).

2. Use the identity for \(\tan(A + B)\):

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

3. Substitute \(\tan \alpha = 2 \tan \beta\) into the equation:

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

4. Simplify to get:

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

5. Solve for \(\tan \beta\):

\(3 \tan \beta = 3 - 6 \tan^2 \beta\)

\(6 \tan^2 \beta + 3 \tan \beta - 3 = 0\)

6. Divide by 3:

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

7. Solve the quadratic equation:

\(\tan \beta = \frac{-1 \pm \sqrt{1 + 8}}{4} = \frac{-1 \pm 3}{4}\)

8. Solutions are \(\tan \beta = \frac{1}{2}\) or \(\tan \beta = -1\).

9. For \(\tan \beta = \frac{1}{2}\), \(\beta = 26.6^\circ\) and \(\tan \alpha = 1\), \(\alpha = 45^\circ\).

10. For \(\tan \beta = -1\), \(\beta = 135^\circ\) and \(\tan \alpha = -2\), \(\alpha = 116.6^\circ\).