Solve tan60^circ/sin60^circ+cos60^circ

$$\frac { \tan 60 ^ { \circ } } { \sin 60 ^ { \circ } + \cos 60 ^ { \circ } }$$

Evaluate

$$3-\sqrt{3}\approx 1.267949192$$

Solution Steps

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

$$\frac{\sqrt{3}}{\sin(60)+\cos(60)}$$

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

$$\frac{\sqrt{3}}{\frac{\sqrt{3}}{2}+\cos(60)}$$

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

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

Since \frac{\sqrt{3}}{2} and \frac{1}{2} have the same denominator, add them by adding their numerators.

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

Divide \sqrt{3} by \frac{\sqrt{3}+1}{2} by multiplying \sqrt{3} by the reciprocal of \frac{\sqrt{3}+1}{2}.

$$\frac{\sqrt{3}\times 2}{\sqrt{3}+1}$$

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

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

Consider \left(\sqrt{3}+1\right)\left(\sqrt{3}-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}.

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

Square \sqrt{3}. Square 1.

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

Subtract 1 from 3 to get 2.

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

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

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

The square of \sqrt{3} is 3.

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

Multiply 2 and 3 to get 6.

$$\frac{6-\sqrt{3}\times 2}{2}$$

Multiply -1 and 2 to get -2.

$$\frac{6-2\sqrt{3}}{2}$$

Divide each term of 6-2\sqrt{3} by 2 to get 3-\sqrt{3}.

$$3-\sqrt{3}$$