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

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

Evaluate

$$4$$

Solution Steps

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

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

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

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

Square \sqrt{3}. Square 1.

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

Subtract 1 from 3 to get 2.

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

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

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

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

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

The square of \sqrt{3} is 3.

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

Add 3 and 1 to get 4.

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

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

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

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

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

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

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

Square \sqrt{3}. Square 1.

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

Subtract 1 from 3 to get 2.

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

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

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

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

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

The square of \sqrt{3} is 3.

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

Add 3 and 1 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}.

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

Add 2 and 2 to get 4.

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

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

$$4$$

Factor

$$2^{2}$$