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To solve this trigonometric expression, we need to simplify the left-hand side (LHS) and show that it equals the right-hand side (RHS). Let's go step-by-step:

### Problem Statement
We need to show:
\[
\frac{\sin \theta + 2 \cos \theta}{\cos \theta - 2 \sin \theta} - \frac{\sin \theta - 2 \cos \theta}{\cos \theta + 2 \sin \theta} = \frac{4}{5 \cos^2 \theta - 4}
\]

### Step-by-Step Solution

#### 1. Combine the Fractions on the LHS

To combine the two fractions on the LHS, find a common denominator:

\[
\text{LHS} = \frac{\sin \theta + 2 \cos \theta}{\cos \theta - 2 \sin \theta} - \frac{\sin \theta - 2 \cos \theta}{\cos \theta + 2 \sin \theta}
\]

The common denominator is:
\[
(\cos \theta - 2 \sin \theta)(\cos \theta + 2 \sin \theta)
\]

So, we rewrite the LHS as:

\[
\text{LHS} = \frac{(\sin \theta + 2 \cos \theta)(\cos \theta + 2 \sin \theta) - (\sin \theta - 2 \cos \theta)(\cos \theta - 2 \sin \theta)}{(\cos \theta - 2 \sin \theta)(\cos \theta + 2 \sin \theta)}
\]

#### 2. Expand the Numerator

Let's expand both products in the numerator:

\[
(\sin \theta + 2 \cos \theta)(\cos \theta + 2 \sin \theta) = \sin \theta \cos \theta + 2 \sin^2 \theta + 2 \cos^2 \theta + 4 \cos \theta \sin \theta
\]

\[
(\sin \theta - 2 \cos \theta)(\cos \theta - 2 \sin \theta) = \sin \theta \cos \theta - 2 \sin^2 \theta - 2 \cos^2 \theta + 4 \cos \theta \sin \theta
\]

Subtracting these two expressions:

\[
(\sin \theta + 2 \cos \theta)(\cos \theta + 2 \sin \theta) - (\sin \theta - 2 \cos \theta)(\cos \theta - 2 \sin \theta)
\]

Simplifying:

\[
= (\sin \theta \cos \theta + 2 \sin^2 \theta + 2 \cos^2 \theta + 4 \cos \theta \sin \theta) - (\sin \theta \cos \theta - 2 \sin^2 \theta - 2 \cos^2 \theta + 4 \cos \theta \sin \theta)
\]

#### 3. Simplify the Numerator

Combine like terms:

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

\[
= 2 \sin^2 \theta + 2 \cos^2 \theta + 2 \sin^2 \theta + 2 \cos^2 \theta
\]

\[
= 4 (\sin^2 \theta + \cos^2 \theta)
\]

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

\[
= 4 \times 1 = 4
\]

#### 4. Simplify the Denominator

The denominator is:

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

\[
= \cos^2 \theta - 4 \sin^2 \theta
\]

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

\[
= \cos^2 \theta - 4(1 - \cos^2 \theta) = \cos^2 \theta - 4 + 4 \cos^2 \theta
\]

\[
= 5 \cos^2 \theta - 4
\]

#### 5. Conclusion

Now we have:

\[
\text{LHS} = \frac{4}{5 \cos^2 \theta - 4} = \text{RHS}
\]

Thus, we have shown that:

\[
\frac{\sin \theta + 2 \cos \theta}{\cos \theta - 2 \sin \theta} - \frac{\sin \theta - 2 \cos \theta}{\cos \theta + 2 \sin \theta} = \frac{4}{5 \cos^2 \theta - 4}.
\]

### The proof is complete!

Would you like any further details or have more questions?

### Relative Questions for Further Practice:
1. How can the Pythagorean identity be used to simplify trigonometric expressions?
2. What are some common trigonometric identities and how are they derived?
3. How do you factor trigonometric expressions?
4. Can you solve similar expressions involving tangent and cotangent?
5. How do trigonometric functions behave as angles approach 0 or 90 degrees?

**Tip:** Practice combining fractions with different trigonometric functions to strengthen your algebraic manipulation skills!

Learn how to solve the trigonometric expression (sin theta + 2 cos theta) / (cos theta - 2 sin theta) using algebraic manipulation and the Pythagorean identity. This problem demonstrates the application of trigonometric identities and algebraic skills. Suitable for advanced high school students.

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