How do I manipulate this equation to show they are equal without actually calculating it $\sqrt{3}\cdot \tan(\frac{\pi}{12})+3 = 2\sqrt{3}$ I calculated both sides to be $3.4641016151377544$.
Hopefully, the LaTeX shows up in the title but if not here it is again:
$\sqrt{3}\cdot \tan(\frac{\pi}{12})+3 =  2\sqrt{3}$
I'm cutting 8 congruent shapes out of fabric to build a sphere. Having this equality known will help me construct it more accurately, using a compass instead of a bunch of lengths and angles. I can tell from the calculation that they are equal but my curiosity got the best of me and I am just wondering if the terms can be expressed differently to prove they are equal without calculating the sqrts or the tan.
By the way it totally possible that I forgot some axioms or something and made some assumptions about this shape that aren't true. Here's a pic of my theoretical sketch:

 A: Note that we have
$$\tan(x/2)=\frac{1-\cos(x)}{\sin(x)}\implies \tan(\pi/12)=\frac{1-\cos(\pi/6)}{\sin(\pi/6)}$$
And you can finish now.
A: 
The key thing is to show $\tan \frac{\pi}{12} = 2 - \sqrt 3$.
To do this, note that $\tan \frac {\pi}6 = \frac 1{\sqrt 3}$. This can easily be derived from the special $(1,\sqrt 3, 2)$ right triangle which you can construct by bisecting an equilateral triangle of side $2$.
By the double-angle formula for tangent, $\tan 2x = \frac{2\tan x}{1 - \tan^2x}$
If you let $\tan \frac{\pi}{12} = t$, you can now write:
$\frac{2t}{1-t^2} = \frac 1{\sqrt 3}$.
Try solving that exactly in radicals. The rest should be easy.
EDIT: As requested by user2661923, I have included a more "geometric" approach, with reference to the figure above. The top triangle is equilateral, the bottom one is isosceles, and both triangles exist within a square of side $2$.
A: Applying the formula for $\tan(a+b)$ (which has plenty of proofs, including geometric) to $a=\frac\pi3$ and $b=-\frac\pi4$, we get$$\tan\frac\pi{12}=\frac{\sqrt3-1}{1+\sqrt3}=2-\sqrt3.$$
