Answer
$$\tan15^\circ=2-\sqrt3$$
6 goes with A.
Work Step by Step
$$\tan15^\circ$$
$15^\circ$ can be written as half of the angle $30^\circ$.
$$15^\circ=\frac{30^\circ}{2}$$
Thus, $$\tan15^\circ=\tan\Big(\frac{30^\circ}{2}\Big)$$
We can apply here the Half-Angle Identity for tangent, which states
$$\tan\Big(\frac{A}{2}\Big)=\frac{\sin A}{1+\cos A}$$
with $A=30^\circ$:
$$\tan15^\circ=\frac{\sin30^\circ}{1+\cos30^\circ}$$
$$\tan15^\circ=\frac{\frac{1}{2}}{1+\frac{\sqrt3}{2}}$$
$$\tan15^\circ=\frac{\frac{1}{2}}{\frac{2+\sqrt3}{2}}$$
Two similar denominators would eliminate each other.
$$\tan15^\circ=\frac{1}{2+\sqrt3}$$
Now multiply both numerator and denominator with $(2-\sqrt3)$.
- Numerator: $1\times(2-\sqrt3)=2-\sqrt3$
- Denominator: $(2+\sqrt3)(2-\sqrt3)=4-3=1$ (since $(A-B)(A+B)=A^2-B^2$)
That means,
$$\tan15^\circ=\frac{2-\sqrt3}{1}$$
$$\tan15^\circ=2-\sqrt3$$
6 goes with A.
