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1. $(\frac{2 + \sqrt{3}}{2 - \sqrt{3}} + \frac{2 - \sqrt{3}}{2 + \sqrt{3}} + \frac{\sqrt{3} - 1}{\sqrt{3} + 1})$ Simplifies to :

A. 2 - $\sqrt{3}$

B. 2 + $\sqrt{3}$

C. 16 - $\sqrt{3}$

D. 40 - $\sqrt{3}$

Solution:

(\frac{2 + \sqrt{3}}{2 - \sqrt{3}} + \frac{2 - \sqrt{3}}{2 + \sqrt{3}} + \frac{\sqrt{3} - 1}{\sqrt{3} + 1})

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

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

\begin{aligned} = {14 + \frac{3 + 1 - 2 \sqrt{3}}{2}} \\ = 14 + \frac{2 (2 - \sqrt{3})}{2} \\ = 14 + 2 - \sqrt{3} \\ = 16 - \sqrt{3} \end{aligned}

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