1. We are asked to simplify the expression $$\frac{\sqrt{2 - \sqrt{3}}}{\sqrt{2 + \sqrt{3}}}$$. 2. To simplify a fraction with square roots in numerator and denominator, we can multiply numerator and denominator by the conjugate of the denominator to rationalize it. 3. The conjugate of $$\sqrt{2 + \sqrt{3}}$$ is $$\sqrt{2 - \sqrt{3}}$$. 4. Multiply numerator and denominator by $$\sqrt{2 - \sqrt{3}}$$: $$\frac{\sqrt{2 - \sqrt{3}}}{\sqrt{2 + \sqrt{3}}} \times \frac{\sqrt{2 - \sqrt{3}}}{\sqrt{2 - \sqrt{3}}} = \frac{\left(\sqrt{2 - \sqrt{3}}\right)^2}{\sqrt{2 + \sqrt{3}} \times \sqrt{2 - \sqrt{3}}}$$ 5. Simplify numerator: $$\left(\sqrt{2 - \sqrt{3}}\right)^2 = 2 - \sqrt{3}$$ 6. Simplify denominator using the difference of squares formula: $$\sqrt{2 + \sqrt{3}} \times \sqrt{2 - \sqrt{3}} = \sqrt{(2 + \sqrt{3})(2 - \sqrt{3})} = \sqrt{2^2 - (\sqrt{3})^2} = \sqrt{4 - 3} = \sqrt{1} = 1$$ 7. So the expression simplifies to: $$\frac{2 - \sqrt{3}}{1} = 2 - \sqrt{3}$$ Final answer: $$2 - \sqrt{3}$$