Solve cos(30^circ)/sqrt{1+frac{cot(60^circ){0+tan(30^circ)}}-sin(30^circ)}

$$\frac{ \cos ( 30 ^ { \circ } ) }{ \sqrt{ 1+ \frac{ \cot ( 60 ^ { \circ } ) }{ 0+ \tan ( 30 ^ { \circ } ) } } - \sin ( 30 ^ { \circ } ) }$$

Evaluate

$$\frac{\sqrt{3}+2\sqrt{6}}{7}\approx 0.947290042$$

Solution Steps

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

$$\frac{\frac{\sqrt{3}}{2}}{\sqrt{1+\frac{\cot(60)}{0+\tan(30)}}-\sin(30)}$$

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

$$\frac{\frac{\sqrt{3}}{2}}{\sqrt{1+\frac{\frac{\sqrt{3}}{3}}{0+\tan(30)}}-\sin(30)}$$

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

$$\frac{\frac{\sqrt{3}}{2}}{\sqrt{1+\frac{\frac{\sqrt{3}}{3}}{0+\frac{\sqrt{3}}{3}}}-\sin(30)}$$

Anything plus zero gives itself.

$$\frac{\frac{\sqrt{3}}{2}}{\sqrt{1+\frac{\frac{\sqrt{3}}{3}}{\frac{\sqrt{3}}{3}}}-\sin(30)}$$

Divide \frac{\sqrt{3}}{3} by \frac{\sqrt{3}}{3} by multiplying \frac{\sqrt{3}}{3} by the reciprocal of \frac{\sqrt{3}}{3}.

$$\frac{\frac{\sqrt{3}}{2}}{\sqrt{1+\frac{\sqrt{3}\times 3}{3\sqrt{3}}}-\sin(30)}$$

Cancel out 3\sqrt{3} in both numerator and denominator.

$$\frac{\frac{\sqrt{3}}{2}}{\sqrt{1+1}-\sin(30)}$$

Add 1 and 1 to get 2.

$$\frac{\frac{\sqrt{3}}{2}}{\sqrt{2}-\sin(30)}$$

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

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

Express \frac{\frac{\sqrt{3}}{2}}{\sqrt{2}-\frac{1}{2}} as a single fraction.

$$\frac{\sqrt{3}}{2\left(\sqrt{2}-\frac{1}{2}\right)}$$

Use the distributive property to multiply 2 by \sqrt{2}-\frac{1}{2}.

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

Rationalize the denominator of \frac{\sqrt{3}}{2\sqrt{2}-1} by multiplying numerator and denominator by 2\sqrt{2}+1.

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

Consider \left(2\sqrt{2}-1\right)\left(2\sqrt{2}+1\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

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

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

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

Calculate 2 to the power of 2 and get 4.

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

The square of \sqrt{2} is 2.

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

Multiply 4 and 2 to get 8.

$$\frac{\sqrt{3}\left(2\sqrt{2}+1\right)}{8-1^{2}}$$

Calculate 1 to the power of 2 and get 1.

$$\frac{\sqrt{3}\left(2\sqrt{2}+1\right)}{8-1}$$

Subtract 1 from 8 to get 7.

$$\frac{\sqrt{3}\left(2\sqrt{2}+1\right)}{7}$$

Use the distributive property to multiply \sqrt{3} by 2\sqrt{2}+1.

$$\frac{2\sqrt{3}\sqrt{2}+\sqrt{3}}{7}$$

To multiply \sqrt{3} and \sqrt{2}, multiply the numbers under the square root.

$$\frac{2\sqrt{6}+\sqrt{3}}{7}$$