EPSG:9829

Note: These formulas have been transcribed from EPSG Guidance Note #7-2. Users are encouraged to use that document rather than the text which follows as reference because limitations in the transcription will be avoided.

First calculate the scale factor at natural origin:
for the south pole case
tF  = tan (pi/4 + latF/2) / {[(1 + e sin(latF)) / (1 – e sin(latF))]^(e/2)}

but for the north pole case
tF  = tan (pi/4 - latF/2) * {[(1 + e sin(latF)) / (1 – e sin(latF))]^(e/2)}

then for both cases
mF = cos(latF)  / (1 – e^2 sin^2(latF))^0.5
ko = mF {[(1+e)^(1+e) (1–e)^(1–e)]0.5} / (2  tF)

The forward and reverse conversions then follow the formulae for the
Polar Stereographic (variant A) method:

For the forward conversion from latitude and longitude, for the south pole case

E = FE + rho * sin(lon – lonO)
N = FN + rho * cos(lon – lonO)
where
t = tan(pi/4 + lat/2) / {[(1 + e sin(lat)) / (1 – e sin(lat))]^(e/2)}
rho = 2*a*ko*t / {[(1+e)^(1+e) (1–e)^(1–e)]^0.5}

For the north pole case, 
rho and E are found as for the south pole case but
t  = tan(pi/4 – lat/2) *  {[(1 + e sin(lat)) / (1 – e sin(lat))]^(e/2)}
N = FN – rho * cos(lon – lonO)

For the reverse conversion from easting and northing to latitude and longitude,
lat = chi + (e^2/2 + 5e^4/24 + e^6/12 + 13e^8/360) sin(2 chi) 
+ (7e^4/48 + 29e^6/240 + 811e^8/11520) sin(4 chi)
+ (7e^6/120 +  81e^8/1120) sin(6 chi)  + (4279e^8/161280) sin(8 chi)

where rho'  = [(E-FE)^2  + (N – FN)^2]^0.5
t'   =rho' {[(1+e)^(1+e) * (1– e)^(1-e)]^0.5} / (2 a ko)
and for the south pole case
chi  = 2 atan(t' ) – pi/2 
but for the north pole case
chi  =  pi/2 - 2 atan t'

Then for for both north and south cases if E = FE, lon = lonO
else for the south pole case
lon = lonO + atan2[(E – FE),(N – FN)]
and for the north pole case
lon = lonO + atan2[(E – FE),(FN – N)]
(see implementation notes in preface for atan2 convention)