prove tan(a+b)=(tan a+tanb)/(1-tana*tanb)

First, we'll write the tangent
identity:

tan(a+b) =
sin(a+b)/cos(a+b)

We'll write the formulas for the sine and
cosine of the sum of angles a and b:

sin(a+b) = sina*cosb +
sinb*cosa

cos(a+b) = cosa*cosb -
sina*sinb

We'll substitute sin(a+b) and cos(a+b) by their
formulas:

tan(a+b) = (sina*cosb + sinb*cosa)/(cosa*cosb -
sina*sinb)

We'll factorize by
cosa*cosb:

tan(a+b) =cosa*cosb*[(sina*cosb/cosa*cosb) +
(sinb*cosa/cosa*cosb)]/cosa*cosb*[1 -
(sina*sinb/cosa*cosb)]

We'll simplify and we'll
get:

tan(a+b) = (sina/cos a + sinb/cos b)/(1 - tan a*tan
b)

tan(a+b) = (tan a + tan b)/(1 - tan a*tan
b) q.e.d.