And if you want proof of that last result,
Let OA, OB and OC be chords of unit length such that angle BOA = α, angle COB = β and angle COA = (α + β).
Let X be a point on the chord OB such that angle OXC = 90°; therefore |OX| = cos β and |CX| = sin β.
Let Y and Z respectively be points on the chords OB and OA such that angles CYZ = 180° and CZO = 90°. Then, |CZ| = sin(α + β) and |OZ| = cos(α + β).
The chord CZ can be split into two components, CY and YZ.
CY is the hypotenuse of right-angled triangle CXY. What is the angle CXY?
Angle OCX = 90° - β. Angle OCY = 90° - (α + β). Angle CXY = angle OCX - angle OCY = α.
As we know |CX| = sin β, and that CX is the adjacent side of right-angled triangle CXY, |CX| = |CY|cos α = sin β.
Therefore, |CY| = sin β / cos α.
YZ is the opposite side of triangle OYZ. Angle YOZ = α, so |YZ| = |OY|sin α.
As OY + YX = OX, and |OX| = cos β, |OY| = |OX| - |YX|.
YX is the opposite side of triangle CXY, so |YX| = |CY|sin α = (sin β sin α) / cos α. Therefore |OY| = cos β - (sin β sin α) / cos α.
Then |YZ| = sin α(cos β - (sin β sin α) / cos α) = sin α cos β - (sin β / cos α) sin² α.
|CZ| = |CY| + |YZ| = sin(α + β)
= (sin β / cos α) + sin α cos β - (sin β / cos α) sin² α
= sin α cos β + (sin β / cos α)(1 - sin² α)
= sin α cos β + (sin β / cos α)cos² α
therefore sin (α + β) = sin α cos β + cos α sin β
( The Mock TurtIe ™ --- Thinks you are a cunt, on, Sat 20 Jun 2009, 2:02, archived)

Furthermore,
As |OZ| = cos(α + β), and OZ is the adjacent side of triangle OYZ, |OZ| = |OY|cos α.
= cos α(cos β - (sin β sin α) / cos α).
Therefore = cos(α + β) = cos α cos β - sin α sin β

As tan(α + β) = sin(α + β) / cos(α + β),
tan(α + β) = (sin α cos β + sin β cos α) / (cos α cos β - sin α sin β)
Dividing through by cos α cos β,
tan(α + β) = ((sin α cos β / cos α cos β) + (sin β cos α / cos α cos β)) / ((cos α cos β / cos α cos β) - (sin α sin β / cos α cos β))
= ((sin α / cos α) + (sin β / cos β)) / (1 - (sin α sin β) / (cos α cos β))
so tan(α + β) = (tan α + tan β) / (1 - tan α tan β)
( The Mock TurtIe ™ --- Thinks you are a cunt, on, Sat 20 Jun 2009, 2:03, archived)