Answer
$$\frac{1}{(1-\sqrt{3})^2}=\frac{2+\sqrt{3}}{2}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{1}{(1-\sqrt{3})^2}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{1}{1-\sqrt{3}-\sqrt{3}+3} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{1}{4-2\sqrt{3}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{1}{4-2\sqrt{3}}\frac{4+2\sqrt{3}}{4+2\sqrt{3}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{4+2\sqrt{3}}{16+8\sqrt{3}-8\sqrt{3}-12} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}\frac{4+2\sqrt{3}}{4} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle6}{\textcircled {6}} } }}}\frac{2+\sqrt{3}}{2}\end{aligned} $$ | |
| ① | $$ (1-\sqrt{3})^2 = \left( 1- \sqrt{3} \right) \cdot \left( 1- \sqrt{3} \right) = 1- \sqrt{3}- \sqrt{3} + 3 $$ |
| ② | Simplify numerator and denominator |
| ③ | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ 4 + 2 \sqrt{3}} $$. |
| ④ | Multiply in a numerator. $$ \color{blue}{ 1 } \cdot \left( 4 + 2 \sqrt{3}\right) = \color{blue}{1} \cdot4+\color{blue}{1} \cdot 2 \sqrt{3} = \\ = 4 + 2 \sqrt{3} $$ Simplify denominator. $$ \color{blue}{ \left( 4- 2 \sqrt{3}\right) } \cdot \left( 4 + 2 \sqrt{3}\right) = \color{blue}{4} \cdot4+\color{blue}{4} \cdot 2 \sqrt{3}\color{blue}{- 2 \sqrt{3}} \cdot4\color{blue}{- 2 \sqrt{3}} \cdot 2 \sqrt{3} = \\ = 16 + 8 \sqrt{3}- 8 \sqrt{3}-12 $$ |
| ⑤ | Simplify numerator and denominator |
| ⑥ | Divide both numerator and denominator by 2. |
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