Show that, for any object on the ecliptic,
tan(δ) = sin(α) tan(ε),
where (α, δ) are the object's Right Ascension and declination,
and ε is the obliquity of the ecliptic.

Use the cosine rule again:
cos KX = cos PX cos KP + sin PX sin KP cos P

On the ecliptic, latitude β = 0
So we have
cos 90° = cos(90°-δ) cos(ε)
+ sin(90°-δ) sin(ε) cos(90°+α)
i.e. 0 = sin(δ) cos(ε) - cos(δ) sin(ε) sin(α)

Divide throughout by cos(δ) cos(ε) to get
tan(δ) = tan(ε) sin(α)

Back to "ecliptic coordinates".