Solve sin^{2}(60^{circ})-cos^2(30^circ)+tan^2(30^circ)

$$\sin ^ { 2 } ( 60 ^ { \circ } ) - \cos ^ { 2 } ( 30 ^ { \circ } ) + \tan ^ { 2 } ( 30 ^ { \circ } )$$

Evaluate

$$\frac{1}{3}\approx 0.333333333$$

Solution Steps

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

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

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

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

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

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

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

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

The square of \sqrt{3} is 3.

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

Calculate 2 to the power of 2 and get 4.

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

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

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

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

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

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

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

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

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

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 4 and 3^{2} is 36. Multiply \frac{\left(\sqrt{3}\right)^{2}-3}{4} times \frac{9}{9}. Multiply \frac{\left(\sqrt{3}\right)^{2}}{3^{2}} times \frac{4}{4}.

$$\frac{9\left(\left(\sqrt{3}\right)^{2}-3\right)}{36}+\frac{4\left(\sqrt{3}\right)^{2}}{36}$$

Since \frac{9\left(\left(\sqrt{3}\right)^{2}-3\right)}{36} and \frac{4\left(\sqrt{3}\right)^{2}}{36} have the same denominator, add them by adding their numerators.

$$\frac{9\left(\left(\sqrt{3}\right)^{2}-3\right)+4\left(\sqrt{3}\right)^{2}}{36}$$

The square of \sqrt{3} is 3.

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

Subtract 3 from 3 to get 0.

$$\frac{0}{4}+\frac{\left(\sqrt{3}\right)^{2}}{3^{2}}$$

Zero divided by any non-zero number gives zero.

$$0+\frac{\left(\sqrt{3}\right)^{2}}{3^{2}}$$

The square of \sqrt{3} is 3.

$$0+\frac{3}{3^{2}}$$

Calculate 3 to the power of 2 and get 9.

$$0+\frac{3}{9}$$

Reduce the fraction \frac{3}{9} to lowest terms by extracting and canceling out 3.

$$0+\frac{1}{3}$$

Add 0 and \frac{1}{3} to get \frac{1}{3}.

$$\frac{1}{3}$$