Solve 4/3cot^230+3sin^260-2sin60-3/4tan^230

$$\frac { 4 } { 3 } \cot ^ { 2 } 30 + 3 \sin ^ { 2 } 60 - 2 \sin 60 - \frac { 3 } { 4 } \tan ^ { 2 } 30$$

Evaluate

$$6-\sqrt{3}\approx 4.267949192$$

Solution Steps

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

$$\frac{4}{3}\left(\sqrt{3}\right)^{2}+3\left(\sin(60)\right)^{2}-2\sin(60)-\frac{3}{4}\left(\tan(30)\right)^{2}$$

The square of \sqrt{3} is 3.

$$\frac{4}{3}\times 3+3\left(\sin(60)\right)^{2}-2\sin(60)-\frac{3}{4}\left(\tan(30)\right)^{2}$$

Multiply \frac{4}{3} and 3 to get 4.

$$4+3\left(\sin(60)\right)^{2}-2\sin(60)-\frac{3}{4}\left(\tan(30)\right)^{2}$$

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

$$4+3\times \left(\frac{\sqrt{3}}{2}\right)^{2}-2\sin(60)-\frac{3}{4}\left(\tan(30)\right)^{2}$$

To raise \frac{\sqrt{3}}{2} to a power, raise both numerator and denominator to the power and then divide.

$$4+3\times \frac{\left(\sqrt{3}\right)^{2}}{2^{2}}-2\sin(60)-\frac{3}{4}\left(\tan(30)\right)^{2}$$

Express 3\times \frac{\left(\sqrt{3}\right)^{2}}{2^{2}} as a single fraction.

$$4+\frac{3\left(\sqrt{3}\right)^{2}}{2^{2}}-2\sin(60)-\frac{3}{4}\left(\tan(30)\right)^{2}$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 4 times \frac{2^{2}}{2^{2}}.

$$\frac{4\times 2^{2}}{2^{2}}+\frac{3\left(\sqrt{3}\right)^{2}}{2^{2}}-2\sin(60)-\frac{3}{4}\left(\tan(30)\right)^{2}$$

Since \frac{4\times 2^{2}}{2^{2}} and \frac{3\left(\sqrt{3}\right)^{2}}{2^{2}} have the same denominator, add them by adding their numerators.

$$\frac{4\times 2^{2}+3\left(\sqrt{3}\right)^{2}}{2^{2}}-2\sin(60)-\frac{3}{4}\left(\tan(30)\right)^{2}$$

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

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

Cancel out 2 and 2.

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

To add or subtract expressions, expand them to make their denominators the same. Multiply \sqrt{3} times \frac{2^{2}}{2^{2}}.

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

Since \frac{4\times 2^{2}+3\left(\sqrt{3}\right)^{2}}{2^{2}} and \frac{\sqrt{3}\times 2^{2}}{2^{2}} have the same denominator, subtract them by subtracting their numerators.

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

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

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

To raise \frac{\sqrt{3}}{3} to a power, raise both numerator and denominator to the power and then divide.

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

Multiply \frac{3}{4} times \frac{\left(\sqrt{3}\right)^{2}}{3^{2}} by multiplying numerator times numerator and denominator times denominator.

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

Cancel out 3 in both numerator and denominator.

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

The square of \sqrt{3} is 3.

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

Multiply 3 and 4 to get 12.

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

Reduce the fraction \frac{3}{12} to lowest terms by extracting and canceling out 3.

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

To add or subtract expressions, expand them to make their denominators the same. Expand 2^{2}.

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

Since \frac{4\times 2^{2}+3\left(\sqrt{3}\right)^{2}-\sqrt{3}\times 2^{2}}{4} and \frac{1}{4} have the same denominator, subtract them by subtracting their numerators.

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

Calculate 2 to the power of 2 and get 4.

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

Multiply 4 and 4 to get 16.

$$\frac{16+3\left(\sqrt{3}\right)^{2}-\sqrt{3}\times 2^{2}}{2^{2}}-\frac{1}{4}$$

The square of \sqrt{3} is 3.

$$\frac{16+3\times 3-\sqrt{3}\times 2^{2}}{2^{2}}-\frac{1}{4}$$

Multiply 3 and 3 to get 9.

$$\frac{16+9-\sqrt{3}\times 2^{2}}{2^{2}}-\frac{1}{4}$$

Add 16 and 9 to get 25.

$$\frac{25-\sqrt{3}\times 2^{2}}{2^{2}}-\frac{1}{4}$$

Calculate 2 to the power of 2 and get 4.

$$\frac{25-\sqrt{3}\times 4}{2^{2}}-\frac{1}{4}$$

Multiply -1 and 4 to get -4.

$$\frac{25-4\sqrt{3}}{2^{2}}-\frac{1}{4}$$

Calculate 2 to the power of 2 and get 4.

$$\frac{25-4\sqrt{3}}{4}-\frac{1}{4}$$

Since \frac{25-4\sqrt{3}}{4} and \frac{1}{4} have the same denominator, subtract them by subtracting their numerators.

$$\frac{25-4\sqrt{3}-1}{4}$$

Do the calculations in 25-4\sqrt{3}-1.

$$\frac{24-4\sqrt{3}}{4}$$

Divide each term of 24-4\sqrt{3} by 4 to get 6-\sqrt{3}.

$$6-\sqrt{3}$$