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

Combine the fractions over a common denominator:

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

Expand \((1 + \sin \theta)^2\):

\(1 + 2\sin \theta + \sin^2 \theta\).

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

\(1 + 2\sin \theta + \sin^2 \theta + \cos^2 \theta = 2 + 2\sin \theta\).

Substitute back:

\(\frac{2 + 2\sin \theta}{\cos \theta (1 + \sin \theta)}\).

Factor out 2:

\(\frac{2(1 + \sin \theta)}{\cos \theta (1 + \sin \theta)}\).

Cancel \(1 + \sin \theta\):

\(\frac{2}{\cos \theta}\).

This matches the right-hand side, proving the identity.

Let \(s = \sin x\) and \(c = \cos x\).

Then \(\tan x = \frac{s}{c}\).

Substitute into the left-hand side (LHS):

\(\left( \frac{1}{c} - \frac{s}{c} \right)^2 = \left( \frac{1-s}{c} \right)^2\)

\(= \frac{(1-s)(1-s)}{c^2}\)

Using the identity \(c^2 = 1 - s^2\), we have:

\(= \frac{(1-s)(1-s)}{1-s^2}\)

Factor the denominator:

\(= \frac{(1-s)(1-s)}{(1-s)(1+s)}\)

Cancel the common factor \((1-s)\):

\(= \frac{1-s}{1+s}\)

This matches the right-hand side (RHS):

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

Start by combining the fractions:

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

This simplifies to:

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

Combine like terms:

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

Since \(1 - \cos^2 \theta = \sin^2 \theta\), the expression becomes:

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

Start with the left-hand side (LHS):

\((\sin \theta + \cos \theta)(1 - \sin \theta \cos \theta)\)

Expand the expression:

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

Factor out common terms:

\(= \sin \theta(1 - \cos^2 \theta) + \cos \theta(1 - \sin^2 \theta)\)

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

\(= \sin^3 \theta + \cos^3 \theta\)

Thus, the identity is proven: \((\sin \theta + \cos \theta)(1 - \sin \theta \cos \theta) \equiv \sin^3 \theta + \cos^3 \theta\).

Start with the left-hand side (LHS):

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

Replace \(\tan \theta\) with \(\frac{\sin \theta}{\cos \theta}\):

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

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

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

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

Factor the denominator:

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

Cancel \(1 - \sin \theta\) from numerator and denominator:

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

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