Simplify the expression $\frac{\sqrt{3}+1}{\sqrt{3}-1}$.

1. Problem Description

Simplify the expression $\frac{\sqrt{3}+1}{\sqrt{3}-1}$.

2. Solution Steps

To simplify the expression $\frac{\sqrt{3}+1}{\sqrt{3}-1}$, we need to rationalize the denominator. This can be done by multiplying both the numerator and the denominator by the conjugate of the denominator. The conjugate of $\sqrt{3}-1$ is $\sqrt{3}+1$.

Thus, we have:

$\frac{\sqrt{3}+1}{\sqrt{3}-1} = \frac{\sqrt{3}+1}{\sqrt{3}-1} \cdot \frac{\sqrt{3}+1}{\sqrt{3}+1}$

Multiplying the numerators gives:

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

Multiplying the denominators gives:

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

So we have:

$\frac{\sqrt{3}+1}{\sqrt{3}-1} = \frac{4 + 2\sqrt{3}}{2}$

Now, divide both terms in the numerator by 2:

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

3. Final Answer

$2 + \sqrt{3}$