Prove these trigonometric identities.

1. Combine the terms on the Left Hand Side into a single fraction with a common denominator.

$\text{LHS} = \frac{1-\cos x}{\sin x} + \frac{\sin x}{1-\cos x}$
$= \frac{(1-\cos x)^2 + \sin^2 x}{\sin x(1-\cos x)}$

2. Expand the squared binomial in the numerator.

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

3. Simplify the numerator using the Pythagorean identity.

Using the identity $\sin^2 x + \cos^2 x = 1$:
$= \frac{1 - 2\cos x + 1}{\sin x(1-\cos x)}$
$= \frac{2 - 2\cos x}{\sin x(1-\cos x)}$

4. Factorise the numerator and cancel the common factor with the denominator to reach the RHS.

$= \frac{2(1-\cos x)}{\sin x(1-\cos x)}$
$= \frac{2}{\sin x}$
$= \text{RHS \quad (proven)}$

1. Express the right-hand side entirely in terms of $\sin x$ and $\cos x$.

$\text{RHS} = \frac{\tan x - \cot x}{\sec x - \csc x} = \frac{\frac{\sin x}{\cos x} - \frac{\cos x}{\sin x}}{\frac{1}{\cos x} - \frac{1}{\sin x}}$

2. Combine the fractions in both the numerator and the denominator over a common denominator.

$= \frac{ \frac{\sin^2 x - \cos^2 x}{\sin x \cos x} }{ \frac{\sin x - \cos x}{\sin x \cos x} }$

3. Simplify the complex fraction, factorise using the difference of two squares, and cancel the common factor.

$= \frac{\sin^2 x - \cos^2 x}{\sin x - \cos x} \\ = \frac{(\sin x - \cos x)(\sin x + \cos x)}{\sin x - \cos x} \\ = \sin x + \cos x \\ = \text{LHS}$

1. Express the left-hand side in terms of $\cos 2x$ as a single fraction.

$\text{LHS} = \frac{1}{2}(1 + \sec 2x) = \frac{1}{2} \left( 1 + \frac{1}{\cos 2x} \right) = \frac{\cos 2x + 1}{2\cos 2x}$

2. Apply the double angle identity to the numerator to cancel the factor of 2.

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

3. Express the denominator using the double angle identity in terms of sine and cosine.

$= \frac{\cos^2 x}{\cos^2 x - \sin^2 x}$

4. Divide the numerator and denominator by $\sin^2 x$ to obtain the right-hand side.

$\begin{aligned} &= \frac{\frac{\cos^2 x}{\sin^2 x}}{\frac{\cos^2 x}{\sin^2 x} - \frac{\sin^2 x}{\sin^2 x}} \\ &= \frac{\cot^2 x}{\cot^2 x - 1} \\ &= \text{RHS} \end{aligned}$

1. Express the right-hand side in terms of $\sin x$ and $\cos x$.

2. Combine the fractions in the numerator and the denominator.

3. Simplify the complex fraction and apply the Pythagorean identity.

4. Apply the double angle formula to reach the left-hand side.

1. Group the terms in the numerator and the denominator to prepare for the sum-to-product formulae.

$\text{LHS} = \frac{(\sin 3x + \sin x) - \sin 2x}{(\cos 3x + \cos x) - \cos 2x}$

2. Apply the sum-to-product (factor) formulae to the grouped terms.

$= \frac{2\sin\left(\frac{3x+x}{2}\right)\cos\left(\frac{3x-x}{2}\right) - \sin 2x}{2\cos\left(\frac{3x+x}{2}\right)\cos\left(\frac{3x-x}{2}\right) - \cos 2x}$
$= \frac{2\sin 2x \cos x - \sin 2x}{2\cos 2x \cos x - \cos 2x}$

3. Factorise the common terms in the numerator and denominator, then simplify to obtain the RHS.

$= \frac{\sin 2x (2\cos x - 1)}{\cos 2x (2\cos x - 1)}$
$= \frac{\sin 2x}{\cos 2x}$
$= \tan 2x \equiv \text{RHS}$