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To determine which of the following choices is equal to the expression:

\[
\frac{\sqrt{6} - \sqrt{2}}{\sqrt{6} + \sqrt{2}}
\]

we'll simplify this expression using the technique of rationalizing the denominator.

### Step-by-Step Solution
1. **Rationalize the Denominator:**
   Multiply both the numerator and the denominator by the conjugate of the denominator, \(\sqrt{6} - \sqrt{2}\):

   \[
   \frac{\sqrt{6} - \sqrt{2}}{\sqrt{6} + \sqrt{2}} \cdot \frac{\sqrt{6} - \sqrt{2}}{\sqrt{6} - \sqrt{2}} = \frac{(\sqrt{6} - \sqrt{2})^2}{(\sqrt{6})^2 - (\sqrt{2})^2}
   \]

2. **Simplify the Denominator:**
   The denominator is a difference of squares:

   \[
   (\sqrt{6})^2 - (\sqrt{2})^2 = 6 - 2 = 4
   \]

   So the expression becomes:

   \[
   \frac{(\sqrt{6} - \sqrt{2})^2}{4}
   \]

3. **Expand the Numerator:**
   Now expand \((\sqrt{6} - \sqrt{2})^2\):

   \[
   (\sqrt{6} - \sqrt{2})^2 = (\sqrt{6})^2 - 2 \cdot \sqrt{6} \cdot \sqrt{2} + (\sqrt{2})^2 = 6 - 2\sqrt{12} + 2
   \]

   Since \(\sqrt{12} = 2\sqrt{3}\), we can substitute:

   \[
   6 - 2\sqrt{12} + 2 = 6 - 4\sqrt{3} + 2 = 8 - 4\sqrt{3}
   \]

   So, the expression now becomes:

   \[
   \frac{8 - 4\sqrt{3}}{4} = 2 - \sqrt{3}
   \]

### Final Answer
The expression simplifies to \(2 - \sqrt{3}\), which matches the answer choice:

\[
\boxed{2 - \sqrt{3}}
\]

Let me know if you would like further details on this solution or if you have any questions!

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### Related Questions
1. What is the process of rationalizing the denominator?
2. How does the difference of squares simplify an expression?
3. Why is the conjugate used in simplifying radical expressions?
4. Can this technique of rationalization be applied to complex numbers?
5. What are other ways to simplify radical expressions?

### Tip
When simplifying an expression with square roots in the denominator, use the conjugate to eliminate the radicals for easier computation.

Which of the following is equal to \(\frac{\sqrt{6} - \sqrt{2}}{\sqrt{6} + \sqrt{2}}\)?

Learn how to simplify the expression \(\frac{\sqrt{6} - \sqrt{2}}{\sqrt{6} + \sqrt{2}}\) by rationalizing the denominator. This problem covers algebraic concepts such as radical expressions, the difference of squares, and rationalization techniques. Suitable for high school students in Grades 9-12.

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