(i) Start with the left-hand side: \(\tan^2 \theta - \sin^2 \theta\).

Using \(\tan \theta = \frac{\sin \theta}{\cos \theta}\), we have \(\tan^2 \theta = \frac{\sin^2 \theta}{\cos^2 \theta}\).

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

Factor out \(\sin^2 \theta\):

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

Using \(1 - \cos^2 \theta = \sin^2 \theta\), we have:

\(\sin^2 \theta \left( \frac{1 - \cos^2 \theta}{\cos^2 \theta} \right) = \sin^2 \theta \left( \frac{\sin^2 \theta}{\cos^2 \theta} \right) = \tan^2 \theta \sin^2 \theta\).

Thus, \(\tan^2 \theta - \sin^2 \theta \equiv \tan^2 \theta \sin^2 \theta\).

(ii) From the identity, \(\tan^2 \theta - \sin^2 \theta = \tan^2 \theta \sin^2 \theta\), the right-hand side is positive for \(0^\circ < \theta < 90^\circ\) because \(\tan^2 \theta\) and \(\sin^2 \theta\) are positive.

Therefore, \(\tan^2 \theta > \sin^2 \theta\), implying \(\tan \theta > \sin \theta\) for \(0^\circ < \theta < 90^\circ\).

(i) Given \(f(0) = 10\), we have:

\(a + b = 10\)

Given \(f\left( \frac{2}{3}\pi \right) = 1\), we have:

\(a - \frac{b}{2} = 1\)

Solving these equations:

From \(a + b = 10\), we get \(a = 10 - b\).

Substitute into \(a - \frac{b}{2} = 1\):

\(10 - b - \frac{b}{2} = 1\)

\(10 - \frac{3b}{2} = 1\)

\(\frac{3b}{2} = 9\)

\(b = 6\)

\(a = 10 - 6 = 4\)

Thus, \(a = 4\) and \(b = 6\).

(ii) The range of \(f(x) = a + b \cos x\) is from \(a - b\) to \(a + b\).

\(a - b = 4 - 6 = -2\)

\(a + b = 4 + 6 = 10\)

Therefore, the range is \(-2\) to \(10\).

(iii) To find \(f\left( \frac{5}{6}\pi \right)\):

\(\cos\left( \frac{5}{6}\pi \right) = -\cos\left( \frac{1}{6}\pi \right) = -\frac{\sqrt{3}}{2}\)

\(f\left( \frac{5}{6}\pi \right) = 4 + 6\left(-\frac{\sqrt{3}}{2}\right)\)

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

(a) Start with the equation \(3 \cos \theta = 8 \tan \theta\).

Since \(\tan \theta = \frac{\sin \theta}{\cos \theta}\), substitute to get:

\(3 \cos \theta = 8 \frac{\sin \theta}{\cos \theta}\)

Multiply through by \(\cos \theta\) to eliminate the fraction:

\(3 \cos^2 \theta = 8 \sin \theta\)

Use the identity \(\cos^2 \theta = 1 - \sin^2 \theta\):

\(3(1 - \sin^2 \theta) = 8 \sin \theta\)

Expand and rearrange to form a quadratic equation:

\(3 - 3 \sin^2 \theta = 8 \sin \theta\)

\(3 \sin^2 \theta + 8 \sin \theta - 3 = 0\)

(b) Factor the quadratic equation:

\((3 \sin \theta - 1)(\sin \theta + 3) = 0\)

Solving \(3 \sin \theta - 1 = 0\) gives:

\(\sin \theta = \frac{1}{3}\)

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

\(\theta = \sin^{-1}\left(\frac{1}{3}\right)\)

\(\theta \approx 19.5^\circ\)

(i) Start with \(f(x) = 2 \sin^2 x - 3 \cos^2 x\).

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

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

Simplify to \(f(x) = 2 - 2 \cos^2 x - 3 \cos^2 x = 2 - 5 \cos^2 x\).

Thus, \(a = 2\) and \(b = -5\).

(ii) The expression \(f(x) = 2 - 5 \cos^2 x\) has its maximum value when \(\cos^2 x = 0\), giving \(f(x) = 2\).

The minimum value occurs when \(\cos^2 x = 1\), giving \(f(x) = 2 - 5 = -3\).

(iii) Solve \(f(x) + 1 = 0\):

\(2 - 5 \cos^2 x + 1 = 0\) simplifies to \(2 - 5 \cos^2 x = -1\).

Thus, \(\cos^2 x = 0.6\).

Solving for \(x\), we find \(x = \cos^{-1}(\pm \sqrt{0.6})\).

This gives \(x \approx 0.685\) and \(x \approx 2.46\).

(i) Using the identity \(\tan(\pi - x) = -\tan x\), we have:

\(\tan(\pi - x) = -k\).

(ii) Using the identity \(\tan\left(\frac{\pi}{2} - x\right) = \cot x = \frac{1}{\tan x}\), we have:

\(\tan\left(\frac{\pi}{2} - x\right) = \frac{1}{k}\).

(iii) Using the identity \(\sin x = \frac{\tan x}{\sqrt{1 + \tan^2 x}}\), we have:

\(\sin x = \frac{k}{\sqrt{1 + k^2}}\).