tan(A + B) and tan(A - B)

Divide sin(A + B) by cos(A + B), then divide top and bottom by cos A cos B, and the tangent formulas come out in tangents alone:

$$\tan\left(A+B\right) = \frac{\tan\left(A\right)+\tan\left(B\right)}{1-\tan\left(A\right)\tan\left(B\right)}$$

$$\tan\left(A-B\right) = \frac{\tan\left(A\right)-\tan\left(B\right)}{1+\tan\left(A\right)\tan\left(B\right)}$$

They need tan A and tan B to exist (neither angle is 90° plus a whole number of half turns) and the bottom not to be 0. On top the sign follows the angle; below it is the opposite. For 75° as 45° + 30°, with tan 45° = 1 and $\tan\left(30^{\circ}\right) = \frac{\sqrt{3}}{3}$:

$$\tan\left(75^{\circ}\right) = \frac{1+\frac{\sqrt{3}}{3}}{1-\frac{\sqrt{3}}{3}} = \frac{3+\sqrt{3}}{3-\sqrt{3}}$$

A root below is cleared by multiplying top and bottom by its conjugate, here 3 + √3: the bottom becomes 9 - 3 = 6, the top ${\left(3+\sqrt{3}\right)}^{2} = 12+6\sqrt{3}$, and $\tan\left(75^{\circ}\right) = 2+\sqrt{3}$.