Start by combining the fractions:

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

Expand the numerator:

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

Simplify the numerator:

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

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

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

Start with the left-hand side: \(\tan^2 \theta - \sin^2 \theta\).

Using \(\tan \theta = \frac{\sin \theta}{\cos \theta}\), we have \(\tan^2 \theta = \frac{\sin^2 \theta}{\cos^2 \theta}\).

Substitute into the expression: \(\frac{\sin^2 \theta}{\cos^2 \theta} - \sin^2 \theta\).

Factor out \(\sin^2 \theta\): \(\sin^2 \theta \left( \frac{1}{\cos^2 \theta} - 1 \right)\).

Using \(1 - \cos^2 \theta = \sin^2 \theta\), rewrite as \(\sin^2 \theta \left( \frac{\sin^2 \theta}{\cos^2 \theta} \right)\).

This simplifies to \(\tan^2 \theta \sin^2 \theta\), which is the right-hand side.

Start with the left-hand side:

\(\frac{\tan \theta}{\sin \theta} - \frac{\sin \theta}{\tan \theta}\)

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

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

Simplify each term:

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

Combine the terms over a common denominator:

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

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

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

Recognize this as \(\tan \theta \sin \theta\):

\(\tan \theta \sin \theta\)

Start with the left-hand side (LHS):

\(\tan x + \frac{1}{\tan x} = \frac{\sin x}{\cos x} + \frac{\cos x}{\sin x}\)

Combine the fractions:

\(\frac{\sin^2 x + \cos^2 x}{\sin x \cos x}\)

Using the Pythagorean identity \(\sin^2 x + \cos^2 x = 1\), the expression becomes:

\(\frac{1}{\sin x \cos x}\)

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

Start with the left-hand side:

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

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

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

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

Factor the denominator:

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

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

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

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