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

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

Evaluate

$$3$$

Short Solution Steps

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

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

The square of \sqrt{3} is 3.

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

To add or subtract expressions, expand them to make their denominators the same. Multiply 1 times \frac{3}{3}.

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

Since \frac{3}{3} and \frac{\sqrt{3}}{3} have the same denominator, subtract them by subtracting their numerators.

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

Divide 1 by \frac{3-\sqrt{3}}{3} by multiplying 1 by the reciprocal of \frac{3-\sqrt{3}}{3}.

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

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

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

Consider \left(3-\sqrt{3}\right)\left(3+\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}.

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

Square 3. Square \sqrt{3}.

$$\frac{3\left(3+\sqrt{3}\right)}{9-3}+\frac{1}{1+\frac{1}{\sqrt{3}}}$$

Subtract 3 from 9 to get 6.

$$\frac{3\left(3+\sqrt{3}\right)}{6}+\frac{1}{1+\frac{1}{\sqrt{3}}}$$

Divide 3\left(3+\sqrt{3}\right) by 6 to get \frac{1}{2}\left(3+\sqrt{3}\right).

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

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

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

The square of \sqrt{3} is 3.

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

To add or subtract expressions, expand them to make their denominators the same. Multiply 1 times \frac{3}{3}.

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

Since \frac{3}{3} and \frac{\sqrt{3}}{3} have the same denominator, add them by adding their numerators.

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

Divide 1 by \frac{3+\sqrt{3}}{3} by multiplying 1 by the reciprocal of \frac{3+\sqrt{3}}{3}.

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

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

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

Consider \left(3+\sqrt{3}\right)\left(3-\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}.

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

Square 3. Square \sqrt{3}.

$$\frac{1}{2}\left(3+\sqrt{3}\right)+\frac{3\left(3-\sqrt{3}\right)}{9-3}$$

Subtract 3 from 9 to get 6.

$$\frac{1}{2}\left(3+\sqrt{3}\right)+\frac{3\left(3-\sqrt{3}\right)}{6}$$

Divide 3\left(3-\sqrt{3}\right) by 6 to get \frac{1}{2}\left(3-\sqrt{3}\right).

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

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

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

Multiply \frac{1}{2} and 3 to get \frac{3}{2}.

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

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

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

Multiply \frac{1}{2} and 3 to get \frac{3}{2}.

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

Multiply \frac{1}{2} and -1 to get -\frac{1}{2}.

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

Since \frac{3}{2} and \frac{3}{2} have the same denominator, add them by adding their numerators.

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

Add 3 and 3 to get 6.

$$\frac{6}{2}+\frac{1}{2}\sqrt{3}-\frac{1}{2}\sqrt{3}$$

Divide 6 by 2 to get 3.

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

Combine \frac{1}{2}\sqrt{3} and -\frac{1}{2}\sqrt{3} to get 0.

$$3$$

Factor

$$3$$