(i) To prove the identity \(\tan(45^\circ + x) + \tan(45^\circ - x) \equiv 2 \sec 2x\):

Use the tangent addition and subtraction formulas:

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

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

Add the two expressions:

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

Combine into a single fraction:

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

Simplify the numerator:

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

Using the identity \(1 + \tan^2 x = \sec^2 x\), we have:

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

Thus, the identity is proven.

(ii) To sketch the graph of \(y = \tan(45^\circ + x) + \tan(45^\circ - x)\) for \(0^\circ \leq x \leq 90^\circ\):

The function has an asymptote at \(x = 45^\circ\) because \(\tan(45^\circ + x)\) and \(\tan(45^\circ - x)\) become undefined.

The graph consists of two branches approaching the asymptote at \(x = 45^\circ\).

1. Use the angle addition and subtraction formulas:

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

2. Apply the formulas:

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

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

3. Substitute into the equation:

\(\frac{\tan \theta + 1}{1 - \tan \theta} - 2 \left(\frac{\tan \theta - 1}{1 + \tan \theta}\right) = 4\)

4. Simplify and solve for \(\tan \theta\):

Multiply through by \((1 - \tan \theta)(1 + \tan \theta)\):

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

5. Simplify further:

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

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

6. Rearrange to form a quadratic equation:

\(7\tan^2 \theta - 2\tan \theta - 1 = 0\)

7. Solve the quadratic equation using the quadratic formula:

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

where \(a = 7, b = -2, c = -1\).

8. Calculate:

\(\tan \theta = \frac{2 \pm \sqrt{(-2)^2 - 4 \times 7 \times (-1)}}{2 \times 7}\)

\(\tan \theta = \frac{2 \pm \sqrt{4 + 28}}{14}\)

\(\tan \theta = \frac{2 \pm \sqrt{32}}{14}\)

\(\tan \theta = \frac{2 \pm 4\sqrt{2}}{14}\)

\(\tan \theta = \frac{1 \pm 2\sqrt{2}}{7}\)

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

\(\theta = 28.7^\circ\) and \(\theta = 165.4^\circ\).

1. Use the identity for \(\sin(A + 45^\circ)\):

\(\sin A \cos 45^\circ + \cos A \sin 45^\circ = 2\sqrt{2} \cos A\)

2. Simplify using \(\cos 45^\circ = \sin 45^\circ = \frac{\sqrt{2}}{2}\):

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

3. Divide through by \(\cos A\):

\(\tan A \cdot \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} = 2\sqrt{2}\)

4. Solve for \(\tan A\):

\(\tan A = 3\)

5. Use the identity \(\sec^2 B = 1 + \tan^2 B\) in the second equation:

\(4(1 + \tan^2 B) + 5 = 12 \tan B\)

6. Simplify and solve the quadratic equation:

\(4\tan^2 B - 12\tan B + 9 = 0\)

7. Factor or use the quadratic formula to find:

\(\tan B = \frac{3}{2}\)

8. Use the identity for \(\tan(A - B)\):

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

9. Substitute \(\tan A = 3\) and \(\tan B = \frac{3}{2}\):

\(\tan(A - B) = \frac{3 - \frac{3}{2}}{1 + 3 \cdot \frac{3}{2}}\)

10. Simplify:

\(\tan(A - B) = \frac{\frac{3}{2}}{1 + \frac{9}{2}} = \frac{\frac{3}{2}}{\frac{11}{2}} = \frac{3}{11}\)

Given:

\(\tan(\theta - \phi) = 3\)

\(\tan \theta + \tan \phi = 1\)

Using the tangent subtraction formula:

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

Substitute \(\tan \theta + \tan \phi = 1\) into the equation:

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

Simplify:

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

Cross-multiply and simplify:

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

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

Rearrange to form a quadratic equation:

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

Solving this quadratic equation gives:

\(\tan \theta = 1.5 \quad \text{or} \quad \tan \theta = -\frac{4}{3}\)

For \(\tan \theta = 1.5\), \(\theta \approx 56.3^\circ\) or \(236.3^\circ\) (only \(56.3^\circ\) is valid).

For \(\tan \theta = -\frac{4}{3}\), \(\theta \approx 126.9^\circ\) or \(306.9^\circ\) (only \(126.9^\circ\) is valid).

Using \(\tan \theta + \tan \phi = 1\), find \(\phi\):

For \(\theta = 56.3^\circ\), \(\phi \approx 161.6^\circ\).

For \(\theta = 126.9^\circ\), \(\phi \approx 63.4^\circ\).

Thus, the possible pairs are \((\theta, \phi) = (135^\circ, 63.4^\circ)\) and \((53.1^\circ, 161.6^\circ)\).

(i) Use the sum-to-product identities:

\(\cos(\theta - 60^\circ) + \cos(\theta + 60^\circ) = 2 \cos \left( \frac{2\theta}{2} \right) \cos \left( \frac{-120^\circ}{2} \right)\)

\(= 2 \cos \theta \cos(-60^\circ)\)

Since \(\cos(-60^\circ) = \cos(60^\circ) = \frac{1}{2}\), we have:

\(2 \cos \theta \cdot \frac{1}{2} = \cos \theta\)

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

(ii) Start with:

\(\frac{\cos(2x - 60^\circ) + \cos(2x + 60^\circ)}{\cos(x - 60^\circ) + \cos(x + 60^\circ)} = 3\)

Using the identity \(\cos A + \cos B = 2 \cos \left( \frac{A+B}{2} \right) \cos \left( \frac{A-B}{2} \right)\), we have:

\(\cos(2x - 60^\circ) + \cos(2x + 60^\circ) = 2 \cos(2x) \cos(60^\circ) = \cos(2x)\)

\(\cos(x - 60^\circ) + \cos(x + 60^\circ) = 2 \cos(x) \cos(60^\circ) = \cos(x)\)

Thus, \(\frac{\cos(2x)}{\cos(x)} = 3\), implying \(\cos(2x) = 3 \cos(x)\).

Using the double angle identity \(\cos(2x) = 2 \cos^2(x) - 1\), we get:

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

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

Solving this quadratic equation for \(\cos(x)\):

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

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

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

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

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

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

Since \(\cos(x)\) must be between -1 and 1, the valid solution is:

\(\cos(x) = \frac{1}{4}(3 - \sqrt{17})\)