Solve frac{1-sqrt{3}}{1+sqrt{3}}1/1

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

Evaluate

$$\sqrt{3}-2\approx -0.267949192$$

Solution Steps

Divide 1 by 1 to get 1.

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

Rationalize the denominator of \frac{1-\sqrt{3}}{1+\sqrt{3}} by multiplying numerator and denominator by 1-\sqrt{3}.

$$\frac{\left(1-\sqrt{3}\right)\left(1-\sqrt{3}\right)}{\left(1+\sqrt{3}\right)\left(1-\sqrt{3}\right)}\times 1$$

Consider \left(1+\sqrt{3}\right)\left(1-\sqrt{3}\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$\frac{\left(1-\sqrt{3}\right)\left(1-\sqrt{3}\right)}{1^{2}-\left(\sqrt{3}\right)^{2}}\times 1$$

Square 1. Square \sqrt{3}.

$$\frac{\left(1-\sqrt{3}\right)\left(1-\sqrt{3}\right)}{1-3}\times 1$$

Subtract 3 from 1 to get -2.

$$\frac{\left(1-\sqrt{3}\right)\left(1-\sqrt{3}\right)}{-2}\times 1$$

Multiply 1-\sqrt{3} and 1-\sqrt{3} to get \left(1-\sqrt{3}\right)^{2}.

$$\frac{\left(1-\sqrt{3}\right)^{2}}{-2}\times 1$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(1-\sqrt{3}\right)^{2}.

$$\frac{1-2\sqrt{3}+\left(\sqrt{3}\right)^{2}}{-2}\times 1$$

The square of \sqrt{3} is 3.

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

Add 1 and 3 to get 4.

$$\frac{4-2\sqrt{3}}{-2}\times 1$$

Divide each term of 4-2\sqrt{3} by -2 to get -2+\sqrt{3}.

$$\left(-2+\sqrt{3}\right)\times 1$$

Use the distributive property to multiply -2+\sqrt{3} by 1.

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