Solve sin60^circ-1/tan60^circ-2tan45^circ

$$\frac { \sin 60 ^ { \circ } - 1 } { \tan 60 ^ { \circ } - 2 \tan 45 ^ { \circ } }$$

Evaluate

$$\frac{1}{2}=0.5$$

Solution Steps

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

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

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

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

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

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

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

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

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

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

Multiply 2 and 1 to get 2.

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

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

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

Cancel out \sqrt{3}-2 in both numerator and denominator.

$$\frac{1}{2}$$