### Probably worthless

Let L be latitude
(measured
north/south
as
positive/negative)
and Z be angle
from zenith
along the meridian
for an astronomical object
on the meridian
(measured
north/south
as
positive/negative),
then obviously (when you draw a
cross section
diagram
of the Earth)
declination δ is

δ = Z + L
where Z = (±)_{N/S}(90°-A_{N/S}) with upper/lower case
for altitude A_{N/S} measured positive from due north/south.
So
δ = (±)_{N/S}(90°-A_{N/S}) + L and A_{N/S} = 90°+(±)_{N/S}(L-δ) and L = δ + (±)_{N/S}(A_{N/S}-90°)
The formula for the altitude of the NCP/SCP is
A_{N/S} = 90°+(±)_{N/S}(L-(±)_{NCP/SCP}90°)
To be explicit:
A_{N}(NCP)=L A_{S}(NCP)=180°-L
A_{S}(SCP)=180°+L A_{S}(SCP)=-L