Start with the left-hand side:

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

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

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

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

Factor the numerator as a difference of squares:

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

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

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

Split the fraction:

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

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

We start with the left-hand side (LHS):

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

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

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

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

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

\(= \tan^2 x \sin^2 x\)

Thus, the identity \(\tan^2 x - \sin^2 x \equiv \tan^2 x \sin^2 x\) is proven.

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

Using \(\tan x = \frac{\sin x}{\cos x}\), rewrite as:

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

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

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

Factor the numerator as a difference of squares:

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

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

\(\frac{1 + \cos x}{\cos x} = \frac{1}{\cos x} + 1\).

Thus, the identity is proven: \(\frac{\sin x \tan x}{1 - \cos x} \equiv 1 + \frac{1}{\cos x}\).

Start with the left-hand side: \((\sin x + \cos x)(1 - \sin x \cos x)\).

Expand the expression: \(\sin x + \cos x - \sin^2 x \cos x - \cos^2 x \sin x\).

Rearrange terms: \(\sin x + \cos x - \sin^2 x \cos x - \cos^2 x \sin x\).

Use the identities \(\sin^2 x = 1 - \cos^2 x\) and \(\cos^2 x = 1 - \sin^2 x\).

Substitute these into the expression: \(\sin x + \cos x - (1 - \cos^2 x) \cos x - (1 - \sin^2 x) \sin x\).

Simplify: \(\sin x + \cos x - \cos x + \cos^3 x - \sin x + \sin^3 x\).

Combine like terms: \(\sin^3 x + \cos^3 x\).

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

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

Start with the left-hand side:

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

Combine the fractions:

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

Use the identity \(1 - s^2 = c^2\):

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

Since \(\tan x = \frac{s}{c}\), we have:

\(\frac{2s^2}{c^2} = 2 \left( \frac{s}{c} \right)^2 = 2 \tan^2 x\)

Thus, the identity is proven.