I couldn't find this information documented anywhere, so I derived it for myself and am sharing it here.

Comments welcome.

Michael,

Great post!

Geometrically is more easy to see L2=L1*tan(theta/2) which is equivalent with your result.

Thanks, Mario

Using the diagram, here's how I arrived to mine:
L3 = L1 cos(theta)
L4 = L2 cos(90-theta) = L2 sin(theta)
L1 = L3+L4

L3 is the horizontal component of the radius at angle theta.
L4 is the horizontal component of the "upper" L2.
The sum thereof is the length L1.

Thus
L1 = L1cos(theta) + L2sin(theta)

L1-L1cos(theta) = L2 sin(theta)

L2 = L1/sin(theta) - L1cos(theta)/sin(theta)

L2 = L1 (1/sin(theta)-1/tan(theta))

After some though, I think I can see your identity - bisecting theta draws two right angle triangles. From the law of sines, you would then have...

L2/sin(theta/2)=L1/sin(90-theta/2)=L1/cos(theta/2)

L2 = L1 * sin(theta/2)/cos(theta/2) = L1 tan(theta/2)

When you see it, you see it, when you don't, you don't.

Or, even more simple, in one of the right angle triangles you've seen, tan(theta/2)=L2/L1

I think also L1 definition is a little bit confusing, for me is just the bend radius (as length).

About your remark: yes, knowing more math the danger is to get complicated but in the end correctness of the result is what really matters.
Your result is correct and can be simplified as
1/sin(theta)-1/tan(theta)=(1-cos(theta))/sin(theta)=
=[2*sin^2(theta/2)]/[2*sin(theta/2)*cos(theta/2)]=
=sin(theta/2)/cos(theta/2)=
=tan(theta/2)