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:

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

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

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

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

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