Solve sin(60)/cos(45)-sin(30)+(tan(60)+1)-1

$$\frac{ \sin ( 60 ) }{ \cos ( 45 ) - \sin ( 30 ) } +( \tan ( 60 ) +1)-1$$

Evaluate

$$\sqrt{6}+2\sqrt{3}\approx 5.913591358$$

Solution Steps

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

$$\frac{\frac{\sqrt{3}}{2}}{\cos(45)-\sin(30)}+\tan(60)+1-1$$

Get the value of \cos(45) from trigonometric values table.

$$\frac{\frac{\sqrt{3}}{2}}{\frac{\sqrt{2}}{2}-\sin(30)}+\tan(60)+1-1$$

Get the value of \sin(30) from trigonometric values table.

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

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

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

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

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

Cancel out 2 in both numerator and denominator.

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

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

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

Consider \left(\sqrt{2}-1\right)\left(\sqrt{2}+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}.

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

Square \sqrt{2}. Square 1.

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

Subtract 1 from 2 to get 1.

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

Anything divided by one gives itself.

$$\sqrt{3}\left(\sqrt{2}+1\right)+\tan(60)+1-1$$

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

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

Subtract 1 from 1 to get 0.

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

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

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

To multiply \sqrt{3} and \sqrt{2}, multiply the numbers under the square root.

$$\sqrt{6}+\sqrt{3}+\sqrt{3}$$

Combine \sqrt{3} and \sqrt{3} to get 2\sqrt{3}.

$$\sqrt{6}+2\sqrt{3}$$