Solve sin60^circ-1/tan60^circ-2+tan45^circ-sqrt{3}cos30^circ

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

Evaluate

$$\frac{-\sqrt{3}-5}{4}\approx -1.683012702$$

Short Solution Steps

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

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

To add or subtract expressions, expand them to make their denominators the same. Multiply 1 times \frac{2}{2}.

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

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

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

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

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

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

$$\frac{\frac{\sqrt{3}-2}{2}}{\sqrt{3}-2+1}-\sqrt{3}\cos(30)$$

Add -2 and 1 to get -1.

$$\frac{\frac{\sqrt{3}-2}{2}}{\sqrt{3}-1}-\sqrt{3}\cos(30)$$

Express \frac{\frac{\sqrt{3}-2}{2}}{\sqrt{3}-1} as a single fraction.

$$\frac{\sqrt{3}-2}{2\left(\sqrt{3}-1\right)}-\sqrt{3}\cos(30)$$

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

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

Express \sqrt{3}\times \frac{\sqrt{3}}{2} as a single fraction.

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

Multiply \sqrt{3} and \sqrt{3} to get 3.

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

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 2\left(\sqrt{3}-1\right) and 2 is 2\left(\sqrt{3}-1\right). Multiply \frac{3}{2} times \frac{\sqrt{3}-1}{\sqrt{3}-1}.

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

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

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

Do the multiplications in \sqrt{3}-2-3\left(\sqrt{3}-1\right).

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

Do the calculations in \sqrt{3}-2-3\sqrt{3}+3.

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

Expand 2\left(\sqrt{3}-1\right).

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

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

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

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

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

Expand \left(2\sqrt{3}\right)^{2}.

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

Calculate 2 to the power of 2 and get 4.

$$\frac{\left(-2\sqrt{3}+1\right)\left(2\sqrt{3}+2\right)}{4\left(\sqrt{3}\right)^{2}-2^{2}}$$

The square of \sqrt{3} is 3.

$$\frac{\left(-2\sqrt{3}+1\right)\left(2\sqrt{3}+2\right)}{4\times 3-2^{2}}$$

Multiply 4 and 3 to get 12.

$$\frac{\left(-2\sqrt{3}+1\right)\left(2\sqrt{3}+2\right)}{12-2^{2}}$$

Calculate 2 to the power of 2 and get 4.

$$\frac{\left(-2\sqrt{3}+1\right)\left(2\sqrt{3}+2\right)}{12-4}$$

Subtract 4 from 12 to get 8.

$$\frac{\left(-2\sqrt{3}+1\right)\left(2\sqrt{3}+2\right)}{8}$$

Use the distributive property to multiply -2\sqrt{3}+1 by 2\sqrt{3}+2 and combine like terms.

$$\frac{-4\left(\sqrt{3}\right)^{2}-2\sqrt{3}+2}{8}$$

The square of \sqrt{3} is 3.

$$\frac{-4\times 3-2\sqrt{3}+2}{8}$$

Multiply -4 and 3 to get -12.

$$\frac{-12-2\sqrt{3}+2}{8}$$

Add -12 and 2 to get -10.

$$\frac{-10-2\sqrt{3}}{8}$$