I need help with this question.
Here is the method step by step.
$\dfrac{1 + \sqrt{3}}{1 - \sqrt{3}} = \dfrac{1 + \sqrt{3}}{1 - \sqrt{3}} * 1 = \dfrac{1 + \sqrt{3}}{1 - \sqrt{3}} * \dfrac{1 + \sqrt{3}}{1 + \sqrt{3}} =$
$ \dfrac{1^2 + \sqrt{3} + \sqrt{3} + (\sqrt{3})^2}{1^2 + \sqrt{3} - \sqrt{3} - (\sqrt{3})^2} = \dfrac{1 + 2\sqrt{3} + 3}{1 - 3} = -\ \dfrac{2(2 + \sqrt{3})}{2}.$
EDIT: I believe Plato read the question incorrectly so he and I disagree in terms of sign only. Of course it is always possible I misread the problem, which was hard to parse on a tablet.
Because I did manage to misread the question. I hate it when people put those tiny jpegs in instead of writing out the question.
Yes of course Plato is correct.
$\dfrac{1 + \dfrac{1}{\sqrt{3}}}{1 - \dfrac{1}{\sqrt{3}}} = \dfrac{\dfrac{\sqrt{3} + 1}{\cancel{\sqrt{3}}}}{\dfrac{\sqrt{3} - 1}{\cancel{\sqrt{3}}}} = \dfrac{\sqrt{3} + 1}{\sqrt{3} - 1}$.