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

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

Evaluate

$$2$$

Solution Steps

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

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

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}.

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

Square \sqrt{3}. Square 1.

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

Subtract 1 from 3 to get 2.

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

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

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

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

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

The square of \sqrt{3} is 3.

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

Add 3 and 1 to get 4.

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

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

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

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

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

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}.

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

Square \sqrt{3}. Square 1.

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

Subtract 1 from 3 to get 2.

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

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

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

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

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

The square of \sqrt{3} is 3.

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

Add 3 and 1 to get 4.

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

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

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

Add 2 and 2 to get 4.

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

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

$$\sqrt{4}$$

Calculate the square root of 4 and get 2.

$$2$$

Factor

$$2$$