Start with \(\frac{\sin \theta - \cos \theta}{\sin \theta + \cos \theta}\).

Divide the numerator and the denominator by \(\cos \theta\):

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

Let \(t = \tan \theta = \frac{\sin \theta}{\cos \theta}\), then:

\(\frac{t - 1}{t + 1}\).

This matches the right-hand side of the identity.

We start with the expression \(\sin^4 \theta - \cos^4 \theta\).

Using the identity \(a^2 - b^2 = (a-b)(a+b)\), we can rewrite it as:

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

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

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

We know that \(\cos^2 \theta = 1 - \sin^2 \theta\), so:

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

Thus, \(\sin^4 \theta - \cos^4 \theta \equiv 2 \sin^2 \theta - 1\).

Let \(s = \sin x\) and \(c = \cos x\). Then \(\tan x = \frac{s}{c}\).

Substitute \(\tan x = \frac{s}{c}\) into the left-hand side:

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

Simplify the expression:

\(= \frac{s+c}{s^2 + c^2}\)

Since \(s^2 + c^2 = 1\), the expression becomes:

\(= s + c\)

This matches the right-hand side, \(\sin x + \cos x\).

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

The left-hand side (LHS) is:

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

Combine the fractions:

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

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

Substitute \(c^2\) in the expression:

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

Factor out \(s\):

\(\frac{s(s + 1)}{c(1+s)}\)

Cancel \(1+s\) from numerator and denominator:

\(\frac{s}{c}\)

Thus, the LHS simplifies to \(\tan \theta\), proving the identity.

Start with the left-hand side (LHS):

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

Put over a common denominator:

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

Simplify the numerator using the identity \(\sin^2 \theta = 1 - \cos^2 \theta\):

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

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

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

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

Recognize this as \(\frac{1}{\tan \theta}\).