what is the following quotient?\n$\frac{1}{1 + sqrt{3}}$\n$\frac{sqrt{3}}{4}$\n$\frac{1+sqrt{3}}{4}$\n$\frac{…

Explanation:

Step1: Rationalize the denominator

Multiply the fraction $\frac{1}{1 + \sqrt{3}}$ by $\frac{1-\sqrt{3}}{1-\sqrt{3}}$. We get $\frac{1\times(1 - \sqrt{3})}{(1+\sqrt{3})(1 - \sqrt{3})}$.

Step2: Expand the denominator

Use the difference - of - squares formula $(a + b)(a - b)=a^{2}-b^{2}$. Here $a = 1$ and $b=\sqrt{3}$, so $(1+\sqrt{3})(1 - \sqrt{3})=1^{2}-(\sqrt{3})^{2}=1 - 3=-2$. And the numerator is $1-\sqrt{3}$. So the fraction becomes $\frac{1-\sqrt{3}}{-2}=\frac{-1+\sqrt{3}}{2}$.

Answer:

$\frac{-1+\sqrt{3}}{2}$