Solve {l}{sqrt{3}tan30^circ+sqrt{2}cos45^circ}{-2sin60^circ+tan60^circ}

$$\left. \begin{array} { l } { \sqrt { 3 } \tan 30 ^ { \circ } + \sqrt { 2 } \cos 45 ^ { \circ } } \\ { - 2 \sin 60 ^ { \circ } + \tan 60 ^ { \circ } } \end{array} \right.$$

Evaluate

$$2$$

Short Solution Steps

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

$$\sqrt{3}\times \frac{\sqrt{3}}{3}+\sqrt{2}\cos(45)-2\sin(60)+\tan(60)$$

Express \sqrt{3}\times \frac{\sqrt{3}}{3} as a single fraction.

$$\frac{\sqrt{3}\sqrt{3}}{3}+\sqrt{2}\cos(45)-2\sin(60)+\tan(60)$$

Get the value of \cos(45) from trigonometric values table.

$$\frac{\sqrt{3}\sqrt{3}}{3}+\sqrt{2}\times \frac{\sqrt{2}}{2}-2\sin(60)+\tan(60)$$

Express \sqrt{2}\times \frac{\sqrt{2}}{2} as a single fraction.

$$\frac{\sqrt{3}\sqrt{3}}{3}+\frac{\sqrt{2}\sqrt{2}}{2}-2\sin(60)+\tan(60)$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 3 and 2 is 6. Multiply \frac{\sqrt{3}\sqrt{3}}{3} times \frac{2}{2}. Multiply \frac{\sqrt{2}\sqrt{2}}{2} times \frac{3}{3}.

$$\frac{2\sqrt{3}\sqrt{3}}{6}+\frac{3\sqrt{2}\sqrt{2}}{6}-2\sin(60)+\tan(60)$$

Since \frac{2\sqrt{3}\sqrt{3}}{6} and \frac{3\sqrt{2}\sqrt{2}}{6} have the same denominator, add them by adding their numerators.

$$\frac{2\sqrt{3}\sqrt{3}+3\sqrt{2}\sqrt{2}}{6}-2\sin(60)+\tan(60)$$

Do the multiplications in 2\sqrt{3}\sqrt{3}+3\sqrt{2}\sqrt{2}.

$$\frac{6+6}{6}-2\sin(60)+\tan(60)$$

Do the calculations in 6+6.

$$\frac{12}{6}-2\sin(60)+\tan(60)$$

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

$$\frac{12}{6}-2\times \frac{\sqrt{3}}{2}+\tan(60)$$

Cancel out 2 and 2.

$$\frac{12}{6}-\sqrt{3}+\tan(60)$$

To add or subtract expressions, expand them to make their denominators the same. Multiply \sqrt{3} times \frac{6}{6}.

$$\frac{12}{6}-\frac{6\sqrt{3}}{6}+\tan(60)$$

Since \frac{12}{6} and \frac{6\sqrt{3}}{6} have the same denominator, subtract them by subtracting their numerators.

$$\frac{12-6\sqrt{3}}{6}+\tan(60)$$

Divide each term of 12-6\sqrt{3} by 6 to get 2-\sqrt{3}.

$$2-\sqrt{3}+\tan(60)$$

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

$$2-\sqrt{3}+\sqrt{3}$$

Combine -\sqrt{3} and \sqrt{3} to get 0.

$$2$$