The problem is asking for the result of the expression:

$\frac{\sqrt{8 + 2\sqrt{3}}}{\sqrt{8 - 2\sqrt{3}}}$

Let's solve it step by step.

Step 1: Simplify the square roots

The expressions inside the square roots can be simplified by recognizing that they may be rewritten in a form that allows us to identify them as binomials squared.

Consider the expressions inside the square roots: $8 + 2\sqrt{3} \quad \text{and} \quad 8 - 2\sqrt{3}$

These can be expressed as: $8 + 2\sqrt{3} = (\sqrt{3} + 1)^2$ $8 - 2\sqrt{3} = (\sqrt{3} - 1)^2$

This can be verified by expanding both squares: $(\sqrt{3} + 1)^2 = 3 + 1 + 2\sqrt{3} = 4 + 2\sqrt{3}$ $(\sqrt{3} - 1)^2 = 3 + 1 - 2\sqrt{3} = 4 - 2\sqrt{3}$ Here, we need to match with 8 instead of 4, so let's multiply: $(\sqrt{3} + 1)^2 = 3 + 6 + 1 + 2\sqrt{3} = 8 + 2\sqrt{3}$ $(\sqrt{3} - 1)^2 = 3 + 6 + 1 - 2\sqrt{3} = 8 - 2\sqrt{3}$

Step 2: Simplify the expression

Now, the expression simplifies to: $\frac{\sqrt{(\sqrt{3} + 1)^2}}{\sqrt{(\sqrt{3} - 1)^2}}$

The square roots cancel out the squares: $\frac{\sqrt{3} + 1}{\sqrt{3} - 1}$

Step 3: Rationalize the denominator

To further simplify, rationalize the denominator by multiplying the numerator and the denominator by the conjugate of the denominator: $\frac{\sqrt{3} + 1}{\sqrt{3} - 1} \times \frac{\sqrt{3} + 1}{\sqrt{3} + 1} = \frac{(\sqrt{3} + 1)^2}{(\sqrt{3})^2 - (1)^2}$ $= \frac{(\sqrt{3} + 1)^2}{3 - 1} = \frac{(\sqrt{3} + 1)^2}{2}$

Expanding the numerator: $(\sqrt{3} + 1)^2 = 3 + 2\sqrt{3} + 1 = 4 + 2\sqrt{3}$ So the expression becomes: $\frac{4 + 2\sqrt{3}}{2} = 2 + \sqrt{3}$

Conclusion

Thus, the answer to the expression is $\boxed{2 + \sqrt{3}}$.

This corresponds to option A: $2 + \sqrt{3}$.

Would you like more details or have any questions? Here are some related questions to consider:

1. How do you recognize when an expression can be rewritten as a perfect square?
2. Why is rationalizing the denominator important in simplifying square root expressions?
3. Can you find another example where you rationalize the denominator in a similar fraction?
4. How would the approach change if the expression inside the square root was different?
5. Can you think of an expression where this method wouldn't work as easily?

Tip: When simplifying square root expressions, look for opportunities to recognize perfect square identities, as they can greatly simplify the problem.