We are given the equation:

$\frac{\sqrt{3} - 1}{\sqrt{3} + 1} = a + b\sqrt{3}$

We need to find the values of $a$ and $b$.

Solution:

1. Rationalize the denominator: We begin by multiplying both the numerator and denominator by the conjugate of the denominator, which is $\sqrt{3} - 1$. This eliminates the square roots from the denominator.

$\frac{\sqrt{3} - 1}{\sqrt{3} + 1} \cdot \frac{\sqrt{3} - 1}{\sqrt{3} - 1} = \frac{(\sqrt{3} - 1)^2}{(\sqrt{3} + 1)(\sqrt{3} - 1)}$

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

$(\sqrt{3} + 1)(\sqrt{3} - 1) = (\sqrt{3})^2 - (1)^2 = 3 - 1 = 2$

So, the equation becomes:

$\frac{(\sqrt{3} - 1)^2}{2}$

3. Expand the numerator: Now, expand $(\sqrt{3} - 1)^2$:

$(\sqrt{3} - 1)^2 = (\sqrt{3})^2 - 2(\sqrt{3})(1) + (1)^2 = 3 - 2\sqrt{3} + 1 = 4 - 2\sqrt{3}$

Thus, the expression becomes:

$\frac{4 - 2\sqrt{3}}{2} = 2 - \sqrt{3}$

4. Compare with the original form: We now compare this with the form $a + b\sqrt{3}$. From the expression $2 - \sqrt{3}$, we see that:

$a = 2 \quad \text{and} \quad b = -1$

Final Answer:

$a = 2, \quad b = -1$

Would you like more details or have any questions?

Here are some related questions for practice:

1. How do you rationalize a denominator involving square roots?
2. What is the formula for the difference of squares?
3. Can you expand $(\sqrt{2} + 1)^2$?
4. How do you simplify complex fractions with radicals?
5. What are the steps to express an irrational number in terms of $a + b\sqrt{c}$?

Tip: When rationalizing the denominator, always use the conjugate to eliminate the square roots.