Start with the equation:

\(4 \sin \theta + \tan \theta = 0\)

Substitute \(\tan \theta = \frac{\sin \theta}{\cos \theta}\):

\(4 \sin \theta + \frac{\sin \theta}{\cos \theta} = 0\)

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

\(4 \sin \theta \cos \theta + \sin \theta = 0\)

Factor out \(\sin \theta\):

\(\sin \theta (4 \cos \theta + 1) = 0\)

This gives two cases:

1. \(\sin \theta = 0\)

2. \(4 \cos \theta + 1 = 0\)

For \(\sin \theta = 0\), \(\theta = 0^\circ, 180^\circ\), but these are outside the given range.

For \(4 \cos \theta + 1 = 0\):

\(\cos \theta = -\frac{1}{4}\)

Using the inverse cosine function, \(\theta \approx 104.5^\circ\) within the range \(0^\circ < \theta < 180^\circ\).

(a) Start with the left-hand side: \(\frac{1 + \sin \theta}{\cos \theta} + \frac{\cos \theta}{1 + \sin \theta}\).

Combine the fractions: \(\frac{(1 + \sin \theta)^2 + \cos^2 \theta}{\cos \theta (1 + \sin \theta)}\).

Expand the numerator: \((1 + \sin \theta)^2 + \cos^2 \theta = 1 + 2\sin \theta + \sin^2 \theta + \cos^2 \theta\).

Using \(\sin^2 \theta + \cos^2 \theta = 1\), the numerator becomes \(2 + 2\sin \theta\).

Thus, \(\frac{2 + 2\sin \theta}{\cos \theta (1 + \sin \theta)} = \frac{2}{\cos \theta}\), proving the identity.

(b) Given \(\frac{1 + \sin \theta}{\cos \theta} + \frac{\cos \theta}{1 + \sin \theta} = \frac{3}{\sin \theta}\), use the identity from part (a): \(\frac{2}{\cos \theta} = \frac{3}{\sin \theta}\).

Cross-multiply to get \(2\sin \theta = 3\cos \theta\).

Divide both sides by \(\cos \theta\): \(\tan \theta = 1.5\).

Find \(\theta\) using \(\tan^{-1}(1.5)\): \(\theta = 0.983\).

Since \(\tan \theta\) is periodic with period \(\pi\), the second solution is \(\theta = 0.983 + \pi = 4.12\).

Start with the equation:

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

Multiply both sides by the denominator:

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

Expand and simplify:

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

Rearrange terms:

\(2 \tan \theta - \tan \theta - 6 \sin \theta - 3 \sin \theta + 2 - 2 = 0\)

\(\tan \theta - 9 \sin \theta = 0\)

Multiply by \(\cos \theta\):

\(\sin \theta - 9 \sin \theta \cos \theta = 0\)

Factorize:

\(\sin \theta (1 - 9 \cos \theta) = 0\)

Set each factor to zero:

\(\sin \theta = 0\) or \(1 - 9 \cos \theta = 0\)

For \(\sin \theta = 0\), \(\theta = 0^\circ\).

For \(1 - 9 \cos \theta = 0\), \(\cos \theta = \frac{1}{9}\).

Calculate \(\theta\) using \(\cos^{-1}\):

\(\theta \approx 83.6^\circ\).

Thus, \(\theta = 0^\circ\) or \(83.6^\circ\) (only answers in the given range).

Start with the equation \(3 \tan(2x + 1) = 1\).

Divide both sides by 3: \(\tan(2x + 1) = \frac{1}{3}\).

Take the inverse tangent: \(2x + 1 = \tan^{-1}\left(\frac{1}{3}\right)\).

Using a calculator, \(\tan^{-1}\left(\frac{1}{3}\right) \approx 0.322\) radians.

Thus, \(2x + 1 = 0.322 + k\pi\) or \(2x + 1 = -0.339 + k\pi\) for integer \(k\).

Solving for \(x\):

\(2x = 0.322 - 1 + k\pi\) or \(2x = -0.339 - 1 + k\pi\).

\(x = \frac{-0.678 + k\pi}{2}\) or \(x = \frac{-1.339 + k\pi}{2}\).

For the smallest positive \(x\), set \(k = 1\):

\(x = \frac{-0.678 + \pi}{2} \approx 1.23\).

For the next smallest \(x\), set \(k = 2\):

\(x = \frac{-1.339 + 2\pi}{2} \approx 2.80\).

(i) Start with the left-hand side (LHS):

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

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

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

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

\(= \frac{1 - \sin x}{1 + \sin x}\)

This matches the right-hand side (RHS), proving the identity.

(ii) Using the result from part (i), we have:

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

Cross-multiply to get:

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

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

\(2 = 4\sin 2x\)

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

Solving \(\sin 2x = \frac{1}{2}\), we find:

\(2x = \frac{\pi}{6}\) or \(2x = \frac{5\pi}{6}\)

Thus, \(x = \frac{\pi}{12}\) or \(x = \frac{5\pi}{12}\).