सिद्धान्तशिरोमणि: गणिताध्याय (भास्कराचार्य - ग्रहगणित, मध्यमाधिकार व स्पष्टाधिकार सटीक)

384
Here sin ω cos λ is called Sa-thribha-graha-ja-kranti or
the declination of a point whose longitude is 90+λ where λ
is the longitude of M. As we have the formula sin δ =
sin ω sin λ, sine of the declination of such a point is equal
to sin ω sin (90+λ)=sin ω cos λ. When δ is very small
sin PMK may be taken to be sin ω cos λ or what is the
same Sa-thribha-graha-ja-krānti as is formulated by Sūrya-
siddhānta.
It may be doubted how sin PB=sin PK sin 90-λ.
(Ref. fig. 82). Let K′ be the centre of the circle PBD,
Fig. 82
K′ being in the plane of PBD. K′P and K′B are radii
of this circle. Since the arc PB stands for 90-λ PK′B=
90-λ. Draw the H sine of arc PB, which is PB′. Now
K′P=H sin ω, as PK′ is ⊥ar drawn on OK
∴ PB′=PK′ sin PK′B
=H sin ω ✕ sin PK′B = (H sin ω ✕ H sin PK′B) / R =
(H sin ω H cos λ) / R
∴ H sin KL = [(H sin ω H cos λ) / R] ✕ H cos δ =
(H sin ω H cos λ) / (H cos δ) as given.