to show a map is a homeomorphism, one needs to do 3 things:

1. show it is bijective (and thus has an inverse map)

2. show it is continuous

3. show its inverse is continuous

to show the map that sends the open disk D^{2}- ∂D^{2}is bijective, suppose that

tan(πr_{1}/2)cos(θ_{1}) = tan(πr_{2}/2)cos(θ_{2});

tan(πr_{1}/2)sin(θ_{1}) = tan(πr_{2}/2)sin(θ_{2})

since the two sides of these two equations represent the same point in R^{2}, the two points have the same norm, so we have:

√[(tan(πr_{1}/2)cos(θ_{1}))^{2}+ (tan(πr_{1}/2)sin(θ_{1}))^{2}] = √[(tan(πr_{2}/2)cos(θ_{2}))^{2}+ (tan(πr_{1}/2)sin(θ_{2}))^{2}]

and simplifying both sides (using cos^{2}(θ_{1}) + sin^{2}(θ_{1}) = cos^{2}(θ_{2}) + sin^{2}(θ_{2}) = 1):

tan(πr_{1}/2) = tan(πr_{2}/2).

since tan(x) is 1-1 on the interval [0,π/2) (remember r is in [0,1)) taking arctangents of both sides, we get:

πr_{1}/2 = πr_{1}/2, and thus r_{1}= r_{2}.

(technically we should restrict r to (0,1) and map the point (0,0) separately, but this is a minor matter).

thus cos(θ_{1}) = cos(θ_{2}) and sin(θ_{1}) = sin(θ_{2}).

from the former equation, we see that θ_{2}= ±θ_{1}, and the second equation then forces us to choose θ_{2}= θ_{1}(since sin is an odd function).

this shows our map is injective. to show it is surjective, let (x,y) be any point in R^{2}.

let r = (2/π)arctan(√(x^{2}+y^{2}))

if x > 0, y > 0, let θ = arctan(y/x) <---we'll ignore the niceties of the difference of definition and assume (x,y) is in the first quadrant.

if x < 0, y > 0, let θ = arctan(y/x) + π <---you can verify it for the other 3, if you wish. it's boring.

if x < 0, y < 0, let θ = arctan(y/x) + π

if x > 0, y < 0, let θ = arctan(y/x) + 2π

if x = 0, y ≥ 0 let θ = π/2

if x = 0, y < 0 let θ = 3π/2

then (rcos(θ),rsin(θ)) maps to:

([tan(π((2/π)arctan(√(x^{2}+y^{2})))/2]cos(arctan(y/x)), [tan(π((2/π)arctan(√(x^{2}+y^{2})))/2]sin(arctan(y/x))

= (tan(arctan(√(x^{2}+y^{2}))cos(arctan(y/x)), tan(arctan(√(x^{2}+y^{2}))sin(arctan(y/x)))

= (√(x^{2}+y^{2})(x/√(x^{2}+y^{2})), (√(x^{2}+y^{2})(y/√(x^{2}+y^{2})) = (x,y),

so our map is surjective, and thus it is bijective.

it should be clear that this is a continuous map, and that its inverse, given by:

(rcos(θ),rsin(θ)) → ((2/π)arctan(r)cos(θ), (2/π)arctan(r)sin(θ)) is likewise continuous.