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

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

Evaluate

$$-4$$

Solution Steps

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

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

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}.

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

Square 1. Square \sqrt{3}.

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

Subtract 3 from 1 to get -2.

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

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

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

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

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

The square of \sqrt{3} is 3.

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

Add 1 and 3 to get 4.

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

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

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

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

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

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}.

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

Square 1. Square \sqrt{3}.

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

Subtract 3 from 1 to get -2.

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

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

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

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

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

The square of \sqrt{3} is 3.

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

Add 1 and 3 to get 4.

$$-2+\sqrt{3}+\frac{4+2\sqrt{3}}{-2}$$

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

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

Subtract 2 from -2 to get -4.

$$-4+\sqrt{3}-\sqrt{3}$$

Combine \sqrt{3} and -\sqrt{3} to get 0.

$$-4$$

Factor

$$-4$$