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

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

Combine the fractions: \(\frac{(1+c)^2 + s^2}{s(1+c)}\).

Expand the numerator: \((1+c)^2 + s^2 = 1 + 2c + c^2 + s^2\).

Use the identity \(c^2 + s^2 = 1\), so the numerator becomes \(1 + 2c + 1 = 2 + 2c\).

Thus, \(\frac{2 + 2c}{s(1+c)} = \frac{2(1+c)}{s(1+c)}\).

Cancel \(1+c\) from the numerator and denominator: \(\frac{2}{s}\).

This matches the right-hand side: \(\frac{2}{\sin \theta}\).

Start by finding a common denominator for the left-hand side:

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

Expand the numerator:

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

Simplify the numerator:

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

Combine like terms:

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

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

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

The denominator simplifies to:

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

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

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

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

Start with the identity \(\cos^2 x = 1 - \sin^2 x\).

Then, \(\cos^4 x = (\cos^2 x)^2 = (1 - \sin^2 x)^2\).

Expand \((1 - \sin^2 x)^2\):

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

Thus, \(\cos^4 x = 1 - 2\sin^2 x + \sin^4 x\).

Let \(c = \cos \theta\) and \(s = \sin \theta\). The left-hand side (LHS) is:

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

Expanding the numerator:

\((1 + c)^2 = 1 + 2c + c^2\)

\((1 - c)^2 = 1 - 2c + c^2\)

Thus,

\((1 + c)^2 - (1 - c)^2 = 1 + 2c + c^2 - (1 - 2c + c^2) = 4c\)

The denominator is:

\((1 - c)(1 + c) = 1 - c^2\)

So,

\(\frac{4c}{1 - c^2}\)

Since \(1 - c^2 = s^2\), we have:

\(\frac{4c}{s^2}\)

Using \(\tan \theta = \frac{s}{c}\), we get:

\(\frac{4}{s \cdot \frac{s}{c}} = \frac{4}{s^2/c} = \frac{4c}{s^2}\)

Thus, the identity is proven as \(\frac{4}{s \cdot \frac{s}{c}} = \frac{4}{s \tan \theta}\).

Start with the left-hand side:

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

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

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

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

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

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

Factor the denominator:

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

Cancel \((1 - c)\) from numerator and denominator:

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

Thus, the identity is proved:

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