Solve sqrt{1/2}*sqrt{6}-tan45^circ

$$\sqrt { \frac { 1 } { 2 } } \times \sqrt { 6 } - \tan 45 ^ { \circ }$$

Evaluate

$$\sqrt{3}-1\approx 0.732050808$$

Short Solution Steps

Rewrite the square root of the division \sqrt{\frac{1}{2}} as the division of square roots \frac{\sqrt{1}}{\sqrt{2}}.

$$\frac{\sqrt{1}}{\sqrt{2}}\sqrt{6}-\tan(45)$$

Calculate the square root of 1 and get 1.

$$\frac{1}{\sqrt{2}}\sqrt{6}-\tan(45)$$

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

$$\frac{\sqrt{2}}{\left(\sqrt{2}\right)^{2}}\sqrt{6}-\tan(45)$$

The square of \sqrt{2} is 2.

$$\frac{\sqrt{2}}{2}\sqrt{6}-\tan(45)$$

Express \frac{\sqrt{2}}{2}\sqrt{6} as a single fraction.

$$\frac{\sqrt{2}\sqrt{6}}{2}-\tan(45)$$

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

$$\frac{\sqrt{2}\sqrt{6}}{2}-1$$

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

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

Since \frac{\sqrt{2}\sqrt{6}}{2} and \frac{2}{2} have the same denominator, subtract them by subtracting their numerators.

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

Do the multiplications in \sqrt{2}\sqrt{6}-2.

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

Divide each term of 2\sqrt{3}-2 by 2 to get \sqrt{3}-1.

$$\sqrt{3}-1$$