Find the value of $$\frac{1 + \sqrt{3}}{1 - \sqrt{3}} + \frac{1 - \sqrt{3}}{1 + \sqrt{3}} \\$$​

Correct option is A

​Solution:

$$\frac{1 + \sqrt{3}}{1 - \sqrt{3}} + \frac{1 - \sqrt{3}}{1 + \sqrt{3}} \\\text{Step 1: Multiply both terms by the conjugate to simplify:} \\= \frac{(1 + \sqrt{3})^2}{(1 - \sqrt{3})(1 + \sqrt{3})} + \frac{(1 - \sqrt{3})^2}{(1 + \sqrt{3})(1 - \sqrt{3})} \\= \frac{(1 + \sqrt{3})^2 + (1 - \sqrt{3})^2}{(1 - \sqrt{3})(1 + \sqrt{3})} \\= \frac{(1^2 + 2\cdot1\cdot\sqrt{3} + (\sqrt{3})^2) + (1^2 - 2\cdot1\cdot\sqrt{3} + (\sqrt{3})^2)}{1^2 - (\sqrt{3})^2} \\= \frac{(1 + 2\sqrt{3} + 3) + (1 - 2\sqrt{3} + 3)}{1 - 3} \\= \frac{(4 + 2\sqrt{3}) + (4 - 2\sqrt{3})}{-2} \\= \frac{8}{-2} \\= -4$$​​