Answer: C
Let \[ 1 + a = 1 + \frac{\sqrt{3}}{2} = \frac{2 + \sqrt{3}}{2} \] Multiply numerator and denominator by 2: \[ = \frac{4 + 2\sqrt{3}}{4} \] Break it down: \[ = \frac{3 + 1 + 2\sqrt{3}}{4} \] Express as a perfect square: \[ = \frac{(\sqrt{3})^2 + 1^2 + 2\sqrt{3} \cdot 1}{4} = \frac{(\sqrt{3} + 1)^2}{4} \] Therefore, \[ \sqrt{1 + a} = \frac{\sqrt{3} + 1}{2} \] Similarly, \[ \sqrt{1 - a} = \frac{\sqrt{3} - 1}{2} \] Now, \[ \sqrt{1 + a} + \sqrt{1 - a} = \frac{\sqrt{3} + 1}{2} + \frac{\sqrt{3} - 1}{2} \] Combine terms: \[ = \frac{\sqrt{3} + 1 + \sqrt{3} - 1}{2} = \frac{2\sqrt{3}}{2} = \sqrt{3} \] Since \(a = \frac{\sqrt{3}}{2}\), \[ \sqrt{1 + a} + \sqrt{1 - a} = 2a \]
Let \[ 1 + a = 1 + \frac{\sqrt{3}}{2} = \frac{2 + \sqrt{3}}{2} \] Multiply numerator and denominator by 2: \[ = \frac{4 + 2\sqrt{3}}{4} \] Break it down: \[ = \frac{3 + 1 + 2\sqrt{3}}{4} \] Express as a perfect square: \[ = \frac{(\sqrt{3})^2 + 1^2 + 2\sqrt{3} \cdot 1}{4} = \frac{(\sqrt{3} + 1)^2}{4} \] Therefore, \[ \sqrt{1 + a} = \frac{\sqrt{3} + 1}{2} \] Similarly, \[ \sqrt{1 - a} = \frac{\sqrt{3} - 1}{2} \] Now, \[ \sqrt{1 + a} + \sqrt{1 - a} = \frac{\sqrt{3} + 1}{2} + \frac{\sqrt{3} - 1}{2} \] Combine terms: \[ = \frac{\sqrt{3} + 1 + \sqrt{3} - 1}{2} = \frac{2\sqrt{3}}{2} = \sqrt{3} \] Since \(a = \frac{\sqrt{3}}{2}\), \[ \sqrt{1 + a} + \sqrt{1 - a} = 2a \]