How to find by hands $\arctan(\sqrt 3 + 2)$

Solving an example with imaginary units.

$$\theta = \arctan(\sqrt3 + 2)$$ $$\theta = 75^o = 5\pi/12$$

Looking at the result maybe it has something to do with $45^o$ and $30^o$ angles. But how to derive the result by hands, if the only thing i have is $\sqrt3 + 2$

It's easy to find $\angle ABC=75^\circ$.

We will use the half-angle formula for tangent: $$ \tan\frac{\theta}{2} = \frac{1-\cos(\theta)}{\sin(\theta)} . $$ We want to get $2+\sqrt{3}$. Remembering the basic values of sine and cosine, I see that $$ 2+\sqrt3 = \frac{1+\frac{\sqrt{3}}{2}}{\frac12} = \frac{1-\cos\frac{5\pi}{6}}{\sin\frac{5\pi}{6}} = \tan\frac{5\pi}{12} $$ and therefore $$\arctan(2+\sqrt3) = \frac{5\pi}{12}$$