Starting with the left-hand side (LHS):

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

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

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

Using \(\sin^2 \theta + \cos^2 \theta = 1\):

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

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

We start with the left-hand side of the identity:

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

Recognize that \(1 - \sin^2 \theta = \cos^2 \theta\) using the Pythagorean identity.

Substitute to get:

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

Rewrite \(1 - 2 \sin^2 \theta\) as \(\cos^2 \theta - \sin^2 \theta\) using the identity \(\cos^2 \theta = 1 - \sin^2 \theta\).

Thus, the expression becomes:

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

Separate the fraction:

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

This simplifies to:

\(1 - \tan^2 \theta\).

Thus, the identity is proven.

Start with the left-hand side of the identity:

\(\left( \frac{1}{\cos x} - \tan x \right) \left( \frac{1}{\sin x} + 1 \right)\)

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

\(\left( \frac{1}{\cos x} - \frac{\sin x}{\cos x} \right) \left( \frac{1}{\sin x} + 1 \right)\)

Simplify the expression:

\(\left( \frac{1 - \sin x}{\cos x} \right) \left( \frac{1 + \sin x}{\sin x} \right)\)

Multiply the numerators and denominators:

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

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

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

Recognize \(1 - \sin^2 x = \cos^2 x\):

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

Simplify by canceling \(\cos x\):

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

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

Thus, the identity is proven.

Start by combining the fractions over a common denominator:

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

Simplify the numerator:

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

The denominator simplifies using the identity \((1 - \sin \theta)(1 + \sin \theta) = 1 - \sin^2 \theta = \cos^2 \theta\):

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

Thus, the expression simplifies to \(2 \tan^2 \theta\), as required.

Start with the left-hand side:

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

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

\(= \frac{2 \tan \theta}{\sin^2 \theta}\)

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

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

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.

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:

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

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 \(\frac{1}{\tan \theta}\).

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.

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.

Start with the left-hand side (LHS):

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

Combine the fractions over a common denominator:

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

Expand the numerator:

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

Thus, the numerator becomes:

\(1 + 2\sin x + \sin^2 x + \cos^2 x.\)

Using the identity \(\sin^2 x + \cos^2 x = 1\), the numerator simplifies to:

\(2 + 2\sin x.\)

So the expression becomes:

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

Factor out 2 from the numerator:

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

Cancel \(1 + \sin x\) from the numerator and denominator:

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

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

Let \(t = \tan x = \frac{\sin x}{\cos x}\). Then \(\tan^2 x = \frac{\sin^2 x}{\cos^2 x}\).

Substitute \(\tan^2 x\) into the identity:

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

Multiply numerator and denominator by \(\cos^2 x\):

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

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

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

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

\((1 - \sin^2 x) - \sin^2 x = 1 - 2\sin^2 x\)

Thus, the identity is proven.

Start with the left-hand side:

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

Find a common denominator:

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

Simplify the numerator:

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

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

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

Now consider the right-hand side:

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

Replace \(\tan^2 \theta\) with \(\frac{\sin^2 \theta}{\cos^2 \theta}\):

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

Multiply numerator and denominator by \(\cos^2 \theta\):

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

Using \(\sin^2 \theta + \cos^2 \theta = 1\), this simplifies to:

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

Both sides are equal, hence the identity is proven.

Start with the left-hand side:

\(\frac{\sin^3 \theta}{\sin \theta - 1} - \frac{\sin^2 \theta}{1 + \sin \theta}\)

Find a common denominator:

\(\frac{\sin^3 \theta (1 + \sin \theta) - \sin^2 \theta (\sin \theta - 1)}{(\sin \theta - 1)(1 + \sin \theta)}\)

Simplify the numerator:

\(\sin^3 \theta + \sin^4 \theta - \sin^3 \theta + \sin^2 \theta = \sin^4 \theta + \sin^2 \theta\)

Thus, the expression becomes:

\(\frac{\sin^4 \theta + \sin^2 \theta}{(\sin \theta - 1)(1 + \sin \theta)}\)

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

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

Using \(\tan \theta = \frac{\sin \theta}{\cos \theta}\), we have:

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

Therefore, the identity is proven as:

\(-\tan^2 \theta (1 + \sin^2 \theta)\)

Start by obtaining a common denominator for the two fractions:

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

Expand the numerators:

\(5 \sin \theta \cos \theta + 2 \sin^2 \theta + 2 \cos^2 \theta - (5 \sin \theta \cos \theta - 2 \sin^2 \theta - 2 \cos^2 \theta)\)

Simplify the expression:

\(\frac{4(\cos^2 \theta + \sin^2 \theta)}{\cos^2 \theta - 4 \sin^2 \theta}\)

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

\(\frac{4}{\cos^2 \theta - 4 \sin^2 \theta}\)

Recognize that \(\cos^2 \theta - 4 \sin^2 \theta = 5 \cos^2 \theta - 4\) by using \(\cos^2 \theta + \sin^2 \theta = 1\) again:

\(\frac{4}{5 \cos^2 \theta - 4}\)

Start with the given equation:

\(\frac{\tan x + \cos x}{\tan x - \cos x} = k\)

Multiply both sides by \(\tan x - \cos x\):

\(\tan x + \cos x = k(\tan x - \cos x)\)

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

\(\frac{\sin x}{\cos x} + \cos x = k \left( \frac{\sin x}{\cos x} - \cos x \right)\)

Clear the fraction by multiplying through by \(\cos x\):

\(\sin x + \cos^2 x = k(\sin x - \cos^2 x)\)

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

\(\sin x + 1 - \sin^2 x = k(\sin x - (1 - \sin^2 x))\)

Simplify the equation:

\(\sin x + 1 - \sin^2 x = k \sin x - k + k \sin^2 x\)

Rearrange terms to gather like terms on one side:

\(k \sin^2 x + \sin^2 x + k \sin x - \sin x - k - 1 = 0\)

Combine like terms:

\((k+1) \sin^2 x + (k-1) \sin x - (k+1) = 0\)

This matches the required expression.

Start with the given equation:

\(\frac{\tan x + \sin x}{\tan x - \sin x} = k.\)

Express \(\tan x\) as \(\frac{\sin x}{\cos x}\):

\(\frac{\frac{\sin x}{\cos x} + \sin x}{\frac{\sin x}{\cos x} - \sin x} = k.\)

Multiply numerator and denominator by \(\cos x\):

\(\frac{\sin x + \sin x \cos x}{\sin x - \sin x \cos x} = k.\)

Factor out \(\sin x\) from both numerator and denominator:

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

Cancel \(\sin x\) from numerator and denominator:

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

Start with the left-hand side:

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

Combine over a common denominator:

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

Expand the numerators:

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

Simplify the numerator:

\(\frac{4 \\sin x}{1 - \\sin^2 x}\)

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

\(\frac{4 \\sin x}{\\cos^2 x}\)

Rewrite \(\\sin x / \\cos x\) as \(\\tan x\):

\(\frac{4 \\tan x}{\\cos x}\)

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