Solve 1/sin(60)-cos(60)-1/sin(60)+cos(60)

$$\frac{ 1 }{ \sin ( 60 ) - \cos ( 60 ) } - \frac{ 1 }{ \sin ( 60 ) + \cos ( 60 ) }$$

Evaluate

$$2$$

Solution Steps

Get the value of \sin(60) from trigonometric values table.

$$\frac{1}{\frac{\sqrt{3}}{2}-\cos(60)}-\frac{1}{\sin(60)+\cos(60)}$$

Get the value of \cos(60) from trigonometric values table.

$$\frac{1}{\frac{\sqrt{3}}{2}-\frac{1}{2}}-\frac{1}{\sin(60)+\cos(60)}$$

Since \frac{\sqrt{3}}{2} and \frac{1}{2} have the same denominator, subtract them by subtracting their numerators.

$$\frac{1}{\frac{\sqrt{3}-1}{2}}-\frac{1}{\sin(60)+\cos(60)}$$

Divide 1 by \frac{\sqrt{3}-1}{2} by multiplying 1 by the reciprocal of \frac{\sqrt{3}-1}{2}.

$$\frac{2}{\sqrt{3}-1}-\frac{1}{\sin(60)+\cos(60)}$$

Rationalize the denominator of \frac{2}{\sqrt{3}-1} by multiplying numerator and denominator by \sqrt{3}+1.

$$\frac{2\left(\sqrt{3}+1\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}-\frac{1}{\sin(60)+\cos(60)}$$

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}.

$$\frac{2\left(\sqrt{3}+1\right)}{\left(\sqrt{3}\right)^{2}-1^{2}}-\frac{1}{\sin(60)+\cos(60)}$$

Square \sqrt{3}. Square 1.

$$\frac{2\left(\sqrt{3}+1\right)}{3-1}-\frac{1}{\sin(60)+\cos(60)}$$

Subtract 1 from 3 to get 2.

$$\frac{2\left(\sqrt{3}+1\right)}{2}-\frac{1}{\sin(60)+\cos(60)}$$

Cancel out 2 and 2.

$$\sqrt{3}+1-\frac{1}{\sin(60)+\cos(60)}$$

Get the value of \sin(60) from trigonometric values table.

$$\sqrt{3}+1-\frac{1}{\frac{\sqrt{3}}{2}+\cos(60)}$$

Get the value of \cos(60) from trigonometric values table.

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

Since \frac{\sqrt{3}}{2} and \frac{1}{2} have the same denominator, add them by adding their numerators.

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

Divide 1 by \frac{\sqrt{3}+1}{2} by multiplying 1 by the reciprocal of \frac{\sqrt{3}+1}{2}.

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

Rationalize the denominator of \frac{2}{\sqrt{3}+1} by multiplying numerator and denominator by \sqrt{3}-1.

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

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}.

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

Square \sqrt{3}. Square 1.

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

Subtract 1 from 3 to get 2.

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

Cancel out 2 and 2.

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

To find the opposite of \sqrt{3}-1, find the opposite of each term.

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

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

$$1+1$$

Add 1 and 1 to get 2.

$$2$$