基础技能 证明下列三角恒等式。

1. Combine the fractions on the left-hand side over a common denominator.

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

2. Expand the squared binomial in the numerator.

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

3. Apply the Pythagorean identity to simplify the numerator.

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

4. Factorise the numerator and simplify the fraction.

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

1. Start with the Right Hand Side (RHS) and express all terms using sine and cosine.

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

2. Combine the terms in the numerator and the denominator into single fractions 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 by cancelling the common denominator.

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

4. Factorise the numerator using the difference of two squares, cancel the common factor, and conclude the proof.

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

1. Express the LHS in terms of $\cos 2x$ with a common denominator.

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

2. Use the double angle formula to simplify the numerator.

Using $\cos 2x = 2\cos^2 x - 1$:
$\frac{1}{2}\left(\frac{2\cos^2 x - 1 + 1}{\cos 2x}\right) = \frac{\cos^2 x}{\cos 2x}$

3. Expand the denominator using another form of the double angle formula.

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

4. Divide numerator and denominator by $\sin^2 x$ to introduce $\cot x$.

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

5. Simplify to obtain the RHS.

$= \frac{\cot^2 x}{\cot^2 x - 1} = \text{RHS}$

1. Express the tangent and cotangent terms on the right-hand side (RHS) in terms of sine and cosine.

$\text{RHS} = \frac{\tan x + \cot x}{\cot x - \tan x} = \frac{\frac{\sin x}{\cos x} + \frac{\cos x}{\sin x}}{\frac{\cos x}{\sin x} - \frac{\sin x}{\cos x}}$

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

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

3. Simplify the complex fraction and apply standard trigonometric identities.

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

4. Convert to the secant function to match the left-hand side (LHS) and complete the proof.

$= \sec 2x = \text{LHS}$

1. Group the first and third terms in the numerator and denominator, and apply the factor formulae.

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

In the denominator:
$\cos 3x + \cos x = 2 \cos\left(\frac{3x + x}{2}\right) \cos\left(\frac{3x - x}{2}\right) = 2 \cos 2x \cos x$

2. Substitute the products back into the left-hand side (LHS) and factorise.

$\begin{aligned} \text{LHS} &= \frac{(\sin 3x + \sin x) - \sin 2x}{(\cos 3x + \cos x) - \cos 2x} \\ &= \frac{2 \sin 2x \cos x - \sin 2x}{2 \cos 2x \cos x - \cos 2x} \\ &= \frac{\sin 2x(2 \cos x - 1)}{\cos 2x(2 \cos x - 1)} \end{aligned}$

3. Simplify the fraction to prove the identity.

$\begin{aligned} \text{LHS} &= \frac{\sin 2x}{\cos 2x} \\ &= \tan 2x \\ &= \text{RHS} \end{aligned}$