(a) To show the given identity, start by obtaining a common denominator:

\(\frac{(\sin \theta + 2 \cos \theta)(\cos \theta + 2 \sin \theta) - (\sin \theta - 2 \cos \theta)(\cos \theta - 2 \sin \theta)}{(\cos \theta - 2 \sin \theta)(\cos \theta + 2 \sin \theta)}\)

Expanding the numerators:

\(5 \sin \theta \cos \theta + 2 \sin^2 \theta + 2 \cos^2 \theta - (5 \sin \theta \cos \theta - 2 \sin^2 \theta - 2 \cos^2 \theta)\)

\(= 4(\cos^2 \theta + \sin^2 \theta)\)

Using \(\cos^2 \theta + \sin^2 \theta = 1\), the expression simplifies to:

\(\frac{4}{\cos^2 \theta - 4 \sin^2 \theta}\)

\(= \frac{4}{5 \cos^2 \theta - 4}\)

(b) For the equation:

\(\frac{4}{5 \cos^2 \theta - 4} = 5\)

Multiply both sides by \(5 \cos^2 \theta - 4\):

\(4 = 5(5 \cos^2 \theta - 4)\)

\(25 \cos^2 \theta = 24\)

\(\cos^2 \theta = \frac{24}{25}\)

\(\cos \theta = \pm \sqrt{\frac{24}{25}} = \pm 0.9798\)

For \(0^\circ < \theta < 180^\circ\), \(\theta = 11.5^\circ\) or \(168.5^\circ\).

(a) Start with \(\frac{\tan x + \cos x}{\tan x - \cos x} = k\).

Use \(\tan x = \frac{\sin x}{\cos x}\) to rewrite the equation: \(\frac{\frac{\sin x}{\cos x} + \cos x}{\frac{\sin x}{\cos x} - \cos x} = k\).

Clear the fraction: \(\frac{\sin x + \cos^2 x}{\sin x - \cos^2 x} = k\).

Use \(\cos^2 x = 1 - \sin^2 x\) to get \(\frac{\sin x + 1 - \sin^2 x}{\sin x - (1 - \sin^2 x)} = k\).

Simplify: \(\frac{\sin x + 1 - \sin^2 x}{\sin^2 x + \sin x - 1} = k\).

Multiply through by the denominator: \((\sin x + 1 - \sin^2 x) = k(\sin^2 x + \sin x - 1)\).

Expand and simplify: \(\sin x + 1 - \sin^2 x = k\sin^2 x + k\sin x - k\).

Rearrange: \((k + 1)\sin^2 x + (k - 1)\sin x - (k + 1) = 0\).

(b) Solve \(\frac{\tan x + \cos x}{\tan x - \cos x} = 4\).

Using the result from part (a), substitute \(k = 4\): \(5\sin^2 x + 3\sin x - 5 = 0\).

Use the quadratic formula: \(\sin x = \frac{-3 \pm \sqrt{9 + 100}}{10}\).

Calculate: \(\sin x = \frac{-3 \pm 11}{10}\).

Solutions: \(\sin x = 0.8\) or \(\sin x = -1.4\) (discard \(\sin x = -1.4\) as it is not possible).

Find \(x\) for \(\sin x = 0.8\): \(x = 48.1^\circ, 131.9^\circ\).

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

\(2 \cos^2 \theta = 7 \cos \theta - 3\)

Rearrange to form a quadratic equation:

\(2 \cos^2 \theta - 7 \cos \theta + 3 = 0\)

Factor the quadratic equation:

\((2 \cos \theta - 1)(\cos \theta - 3) = 0\)

Set each factor to zero and solve for \(\cos \theta\):

\(2 \cos \theta - 1 = 0\) gives \(\cos \theta = \frac{1}{2}\)

\(\cos \theta - 3 = 0\) gives \(\cos \theta = 3\) (not possible since \(|\cos \theta| \leq 1\))

For \(\cos \theta = \frac{1}{2}\), \(\theta = \pm 60^\circ\) within the given range.

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

(a) Start with the identity \(1 - \tan^2 \theta\).

We know \(\tan^2 \theta = \frac{\sin^2 \theta}{\cos^2 \theta}\), so \(1 - \tan^2 \theta = \frac{\cos^2 \theta - \sin^2 \theta}{\cos^2 \theta}\).

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

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

(b) Given \(\frac{1 - 2 \sin^2 \theta}{1 - \sin^2 \theta} = 2 \tan^4 \theta\), substitute \(1 - \tan^2 \theta = 2 \tan^4 \theta\).

This simplifies to \(1 - \tan^2 \theta = 2 \tan^4 \theta \Rightarrow 2 \tan^4 \theta + \tan^2 \theta - 1 = 0\).

Let \(x = \tan^2 \theta\), then \(2x^2 + x - 1 = 0\).

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

\(x = \frac{-1 \pm \sqrt{1 + 8}}{4} = \frac{-1 \pm 3}{4}\).

Thus, \(x = 0.5\) or \(x = -1\). Since \(x = \tan^2 \theta\), \(x = -1\) is not valid.

So, \(\tan^2 \theta = 0.5\) leading to \(\tan \theta = \pm \sqrt{0.5}\).

\(\theta = 35.3^\circ\) and \(\theta = 144.7^\circ\) (A.W.R.T).

Start with the equation:

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

Multiply through by \(\tan^2 \theta\) to eliminate the fraction:

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

Let \(x = \tan^2 \theta\), then the equation becomes:

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

Factor the quadratic equation:

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

Thus, \(x = \frac{2}{3}\) or \(x = -1\).

Since \(x = \tan^2 \theta\), \(x\) must be non-negative, so \(x = -1\) is not valid.

Therefore, \(\tan^2 \theta = \frac{2}{3}\).

Take the square root to find \(\tan \theta\):

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

Calculate \(\theta\) using the inverse tangent function:

\(\theta = \tan^{-1}(\sqrt{\frac{2}{3}}) \approx 39.2^\circ\)

Since \(\tan \theta\) is positive in the first and third quadrants, the solutions are:

\(\theta = 39.2^\circ\) and \(\theta = 180^\circ - 39.2^\circ = 140.8^\circ\)