Solve 2tan^260^circ/1+tan^230^circ

$$\frac { 2 \tan ^ { 2 } 60 ^ { \circ } } { 1 + \tan ^ { 2 } 30 ^ { \circ } }$$

Evaluate

$$\frac{9}{2}=4.5$$

Solution Steps

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

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

The square of \sqrt{3} is 3.

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

Multiply 2 and 3 to get 6.

$$\frac{6}{1+\left(\tan(30)\right)^{2}}$$

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

$$\frac{6}{1+\left(\frac{\sqrt{3}}{3}\right)^{2}}$$

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

$$\frac{6}{1+\frac{\left(\sqrt{3}\right)^{2}}{3^{2}}}$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 1 times \frac{3^{2}}{3^{2}}.

$$\frac{6}{\frac{3^{2}}{3^{2}}+\frac{\left(\sqrt{3}\right)^{2}}{3^{2}}}$$

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

$$\frac{6}{\frac{3^{2}+\left(\sqrt{3}\right)^{2}}{3^{2}}}$$

Divide 6 by \frac{3^{2}+\left(\sqrt{3}\right)^{2}}{3^{2}} by multiplying 6 by the reciprocal of \frac{3^{2}+\left(\sqrt{3}\right)^{2}}{3^{2}}.

$$\frac{6\times 3^{2}}{3^{2}+\left(\sqrt{3}\right)^{2}}$$

Calculate 3 to the power of 2 and get 9.

$$\frac{6\times 9}{3^{2}+\left(\sqrt{3}\right)^{2}}$$

Multiply 6 and 9 to get 54.

$$\frac{54}{3^{2}+\left(\sqrt{3}\right)^{2}}$$

Calculate 3 to the power of 2 and get 9.

$$\frac{54}{9+\left(\sqrt{3}\right)^{2}}$$

The square of \sqrt{3} is 3.

$$\frac{54}{9+3}$$

Add 9 and 3 to get 12.

$$\frac{54}{12}$$

Reduce the fraction \frac{54}{12} to lowest terms by extracting and canceling out 6.

$$\frac{9}{2}$$