Solve {left(-1/2right)}^-2-2+sqrt{3}+1-3tan(30^circ)-sqrt{-2^2}

$${ \left(- \frac{ 1 }{ 2 } \right) }^{ -2 } -2+ \sqrt{ 3 } +1-3 \tan ( 30 ^ { \circ } ) - \sqrt{ { -2 }^{ 2 } }$$

Evaluate

$$1$$

Solution Steps

Calculate -\frac{1}{2} to the power of -2 and get 4.

$$4-2+\sqrt{3}+1-3\tan(30)-\sqrt{\left(-2\right)^{2}}$$

Subtract 2 from 4 to get 2.

$$2+\sqrt{3}+1-3\tan(30)-\sqrt{\left(-2\right)^{2}}$$

Add 2 and 1 to get 3.

$$3+\sqrt{3}-3\tan(30)-\sqrt{\left(-2\right)^{2}}$$

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

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

Cancel out 3 and 3.

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

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

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

Calculate -2 to the power of 2 and get 4.

$$3-\sqrt{4}$$

Calculate the square root of 4 and get 2.

$$3-2$$

Subtract 2 from 3 to get 1.

$$1$$