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:

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