Solve 1/cos30^circ-frac{1{5}}

$$\frac { 1 } { \cos 30 ^ { \circ } - \frac { 1 } { 5 } }$$

Evaluate

$$\frac{50\sqrt{3}+20}{71}\approx 1.501444231$$

Short Solution Steps

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

$$\frac{1}{\frac{\sqrt{3}}{2}-\frac{1}{5}}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 2 and 5 is 10. Multiply \frac{\sqrt{3}}{2} times \frac{5}{5}. Multiply \frac{1}{5} times \frac{2}{2}.

$$\frac{1}{\frac{5\sqrt{3}}{10}-\frac{2}{10}}$$

Since \frac{5\sqrt{3}}{10} and \frac{2}{10} have the same denominator, subtract them by subtracting their numerators.

$$\frac{1}{\frac{5\sqrt{3}-2}{10}}$$

Divide 1 by \frac{5\sqrt{3}-2}{10} by multiplying 1 by the reciprocal of \frac{5\sqrt{3}-2}{10}.

$$\frac{10}{5\sqrt{3}-2}$$

Rationalize the denominator of \frac{10}{5\sqrt{3}-2} by multiplying numerator and denominator by 5\sqrt{3}+2.

$$\frac{10\left(5\sqrt{3}+2\right)}{\left(5\sqrt{3}-2\right)\left(5\sqrt{3}+2\right)}$$

Consider \left(5\sqrt{3}-2\right)\left(5\sqrt{3}+2\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$\frac{10\left(5\sqrt{3}+2\right)}{\left(5\sqrt{3}\right)^{2}-2^{2}}$$

Expand \left(5\sqrt{3}\right)^{2}.

$$\frac{10\left(5\sqrt{3}+2\right)}{5^{2}\left(\sqrt{3}\right)^{2}-2^{2}}$$

Calculate 5 to the power of 2 and get 25.

$$\frac{10\left(5\sqrt{3}+2\right)}{25\left(\sqrt{3}\right)^{2}-2^{2}}$$

The square of \sqrt{3} is 3.

$$\frac{10\left(5\sqrt{3}+2\right)}{25\times 3-2^{2}}$$

Multiply 25 and 3 to get 75.

$$\frac{10\left(5\sqrt{3}+2\right)}{75-2^{2}}$$

Calculate 2 to the power of 2 and get 4.

$$\frac{10\left(5\sqrt{3}+2\right)}{75-4}$$

Subtract 4 from 75 to get 71.

$$\frac{10\left(5\sqrt{3}+2\right)}{71}$$

Use the distributive property to multiply 10 by 5\sqrt{3}+2.

$$\frac{50\sqrt{3}+20}{71}$$