1. Use the identity \(\cot 2\theta = \frac{1}{\tan 2\theta}\) and \(\tan 2\theta = \frac{2 \tan \theta}{1 - \tan^2 \theta}\) to express \(\cot 2\theta\) in terms of \(\tan \theta\):

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

2. Use the identity \(\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}\) to express \(\tan(\theta + 45^\circ)\):

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

3. Substitute these into the original equation:

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

4. Simplify and rearrange to form a quadratic equation:

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

5. Solve the quadratic equation using the quadratic formula \(\tan \theta = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 2, b = 1, c = -1\):

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

6. This gives \(\tan \theta = \frac{1}{2}\) or \(\tan \theta = -1\).

7. Since \(0^\circ < \theta < 90^\circ\), \(\tan \theta = \frac{1}{2}\) is valid.

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

1. Use the tangent addition and subtraction formulas:

\(\tan(A \pm B) = \frac{\tan A \pm \tan B}{1 \mp \tan A \tan B}\)

2. For \(\tan(\theta + 60^\circ)\):

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

3. For \(\tan(60^\circ - \theta)\):

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

4. Set the equation:

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

5. Clear the fractions by multiplying through by \((1 - \tan \theta \cdot \sqrt{3})(1 + \tan \theta \cdot \sqrt{3})\).

6. Simplify and rearrange to form a quadratic equation:

\(3\tan^2 \theta + 4\tan \theta - 1 = 0\)

7. Solve the quadratic equation using the quadratic formula:

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

where \(a = 3, b = 4, c = -1\).

8. Calculate:

\(\tan \theta = \frac{-4 \pm \sqrt{16 + 12}}{6} = \frac{-4 \pm \sqrt{28}}{6} = \frac{-4 \pm 2\sqrt{7}}{6}\)

9. Simplify:

\(\tan \theta = \frac{-2 \pm \sqrt{7}}{3}\)

10. Find \(\theta\) using \(\tan^{-1}\):

\(\theta \approx 12.1^\circ\) and \(\theta \approx 122.9^\circ\).

(a) Start by expressing the left-hand side as a single fraction:

\(\frac{\cos 3x}{\sin x} + \frac{\sin 3x}{\cos x} = \frac{\cos 3x \cdot \cos x + \sin 3x \cdot \sin x}{\sin x \cdot \cos x}\)

Use the angle addition formula \(\cos(A - B) = \cos A \cos B + \sin A \sin B\):

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

Use the double angle identity \(\sin 2x = 2 \sin x \cos x\):

\(= \frac{\cos 2x}{\frac{1}{2} \sin 2x} = \frac{2 \cos 2x}{\sin 2x} = 2 \cot 2x\)

Thus, the identity is proven.

(b) Set \(\frac{\cos 3x}{\sin x} + \frac{\sin 3x}{\cos x} = 4\):

\(2 \cot 2x = 4\) \(\cot 2x = 2\)

Then, \(\tan 2x = \frac{1}{2}\).

Find \(2x\):

\(2x = \tan^{-1}\left(\frac{1}{2}\right) \approx 0.464 \text{ or } 0.464 + n\pi\)

For \(0 < x < \pi\), solve for \(x\):

\(x = \frac{0.464}{2} \approx 0.232\)

And for the next solution:

\(x = \frac{0.464 + \pi}{2} \approx 1.80\)

Thus, \(x \approx 0.232, 1.80\).

(i) Start by expanding \(\tan(2x + x)\) using the angle addition formula:

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

Using \(\tan 2x = \frac{2 \tan x}{1 - \tan^2 x}\), substitute into the equation:

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

Simplify the expression:

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

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

Given \(\tan 3x = 3 \cot x = \frac{3}{\tan x}\), equate and simplify:

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

Cross-multiply and simplify:

\(3 \tan^2 x - \tan^4 x = 3 - 6 \tan^2 x\)

Rearrange to obtain:

\(\tan^4 x - 12 \tan^2 x + 3 = 0\)

(ii) Solve \(\tan^4 x - 12 \tan^2 x + 3 = 0\) by substituting \(y = \tan^2 x\):

\(y^2 - 12y + 3 = 0\)

Use the quadratic formula:

\(y = \frac{12 \pm \sqrt{144 - 12}}{2}\)

\(y = \frac{12 \pm \sqrt{132}}{2}\)

\(y = 6 \pm \sqrt{33}\)

Since \(y = \tan^2 x\), solve for \(x\):

\(\tan x = \sqrt{6 + \sqrt{33}}\) or \(\tan x = \sqrt{6 - \sqrt{33}}\)

Calculate \(x\) for \(0^\circ < x < 90^\circ\):

\(x = 26.8^\circ\) and \(x = 73.7^\circ\)

1. Start with the equation:

\(\cot \theta - \cot(\theta + 45^\circ) = 3\)

2. Use the identity \(\cot(\theta + 45^\circ) = \frac{1 - \tan \theta}{\tan \theta + 1}\).

3. Substitute \(\cot \theta = \frac{1}{\tan \theta}\) and \(\cot(\theta + 45^\circ) = \frac{1 - \tan \theta}{\tan \theta + 1}\) into the equation:

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

4. Simplify and rearrange to form a quadratic equation:

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

5. Expand and simplify:

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

6. Solve the quadratic equation using the quadratic formula:

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

where \(a = 2\), \(b = 3\), \(c = -1\).

7. Calculate:

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

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

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

8. Find \(\theta\) for \(0^\circ < \theta < 180^\circ\):

\(\theta = 15.7^\circ\) or \(\theta = 119.3^\circ\).