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.