(a) To prove the identity:

Start with \(\frac{1 + \sin x}{1 - \sin x} - \frac{1 - \sin x}{1 + \sin x}\).

Use a common denominator: \((1 - \sin x)(1 + \sin x) = 1 - \sin^2 x = \cos^2 x\).

Numerator becomes \((1 + \sin x)^2 - (1 - \sin x)^2 = 1 + 2\sin x + \sin^2 x - (1 - 2\sin x + \sin^2 x) = 4\sin x\).

Thus, \(\frac{4\sin x}{\cos^2 x} = \frac{4\sin x}{\cos x \cdot \cos x} = \frac{4\tan x}{\cos x}\).

Identity is proven.

(b) Solve \(\frac{4\tan x}{\cos x} = 8\tan x\).

Divide both sides by \(\tan x\) (assuming \(\tan x \neq 0\)): \(\frac{4}{\cos x} = 8\).

\(\cos x = \frac{1}{2}\).

\(x = \frac{\pi}{3}\) within the given range.

Also, consider \(\tan x = 0\) which gives \(x = 0\).

Thus, solutions are \(x = 0\) or \(x = \frac{\pi}{3}\).

Start with the equation:

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

Cross-multiply to get:

\(\tan \theta + 2 \sin \theta = 3(\tan \theta - 2 \sin \theta)\)

Expand the right side:

\(\tan \theta + 2 \sin \theta = 3 \tan \theta - 6 \sin \theta\)

Rearrange terms:

\(2 \tan \theta - 8 \sin \theta = 0\)

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

\(2 \sin \theta (\tan \theta - 4 \cos \theta) = 0\)

Since \(\sin \theta = 0\) is not valid in the given range, solve:

\(\tan \theta = 4 \cos \theta\)

Using \(\tan \theta = \frac{\sin \theta}{\cos \theta}\), we have:

\(\frac{\sin \theta}{\cos \theta} = 4 \cos \theta\)

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

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

\(\sin \theta = 4(1 - \sin^2 \theta)\)

\(\sin \theta = 4 - 4 \sin^2 \theta\)

Rearrange to form a quadratic equation:

\(4 \sin^2 \theta + \sin \theta - 4 = 0\)

Using the quadratic formula, solve for \(\sin \theta\):

\(\sin \theta = \frac{-1 \pm \sqrt{1 + 64}}{8}\)

\(\sin \theta = \frac{-1 \pm 9}{8}\)

\(\sin \theta = 1\) or \(\sin \theta = -\frac{5}{4}\)

Only \(\sin \theta = 1\) is valid, leading to \(\theta = 90^\circ\), but this is not in the solution range.

Thus, solve \(\cos \theta = \frac{1}{4}\):

\(\theta = 75.5^\circ\) only.

(a) Start with the expression \(\left( \frac{1}{\cos x} - \tan x \right) \left( \frac{1}{\sin x} + 1 \right)\).

Use the identity \(\tan x = \frac{\sin x}{\cos x}\) to rewrite \(\tan x\) as \(\frac{\sin x}{\cos x}\).

Substitute to get \(\left( \frac{1}{\cos x} - \frac{\sin x}{\cos x} \right) \left( \frac{1}{\sin x} + 1 \right)\).

Simplify the first term: \(\frac{1 - \sin x}{\cos x}\).

Simplify the second term: \(\frac{1 + \sin x}{\sin x}\).

Multiply the two expressions: \(\frac{(1 - \sin x)(1 + \sin x)}{\cos x \sin x}\).

Use the identity \((1 - \sin x)(1 + \sin x) = 1 - \sin^2 x = \cos^2 x\).

Thus, the expression becomes \(\frac{\cos^2 x}{\cos x \sin x} = \frac{\cos x}{\sin x}\).

Recognize \(\frac{\cos x}{\sin x} = \frac{1}{\tan x}\), proving the identity.

(b) From part (a), we have \(\frac{1}{\tan x} = 2 \tan^2 x\).

Rearrange to get \(\tan^3 x = \frac{1}{2}\).

Take the cube root: \(\tan x = \left( \frac{1}{2} \right)^{1/3}\).

Calculate \(x\) using \(\tan^{-1} \left( \left( \frac{1}{2} \right)^{1/3} \right)\).

Within the range \(0^\circ \leq x \leq 180^\circ\), \(x \approx 38.4^\circ\).

(a) Start by putting the expression over a single common denominator:

\(\frac{\sin \theta (1 + \sin \theta) - \sin \theta (1 - \sin \theta)}{(1 - \sin \theta)(1 + \sin \theta)}\)

This simplifies to:

\(\frac{\sin \theta + \sin^2 \theta - \sin \theta + \sin^2 \theta}{1 - \sin^2 \theta}\)

\(= \frac{2 \sin^2 \theta}{\cos^2 \theta}\)

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

(b) From part (a), we have:

\(2 \tan^2 \theta = 8\)

\(\tan^2 \theta = 4\)

\(\tan \theta = \pm 2\)

Solving for \(\theta\) in the range \(0^\circ < \theta < 180^\circ\), we find:

\(\theta = 63.4^\circ, \ 116.6^\circ\)

(a) Start with the left-hand side:

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

\(= \frac{2 \tan \theta}{\sin^2 \theta}\)

\(= \frac{2 \sin \theta}{\cos \theta \sin^2 \theta}\)

\(= \frac{2}{\sin \theta \cos \theta}\)

Thus, the identity is proven.

(b) Given:

\(\frac{2}{\sin \theta \cos \theta} = \frac{6 \cos \theta}{\sin \theta}\)

\(\Rightarrow \frac{2}{\cos \theta} = 6\)

\(\Rightarrow \cos^2 \theta = \frac{1}{3}\)

\(\Rightarrow \cos \theta = \pm 0.5774\)

For \(0^\circ < \theta < 180^\circ\), \(\theta = 54.7^\circ, 125.3^\circ\).