Spherical Law of Tangents

Theorem

Let $\triangle ABC$ be a spherical triangle on the surface of a sphere whose center is $O$.

Let the sides $a, b, c$ of $\triangle ABC$ be measured by the angles subtended at $O$, where $a, b, c$ are opposite $A, B, C$ respectively.

Then:

$\dfrac {\tan \frac 1 2 \paren {A + B} } {\tan \frac 1 2 \paren {A - B} } = \dfrac {\tan \frac 1 2 \paren {a + b} } {\tan \frac 1 2 \paren {a - b} }$

$\ds \tan \dfrac {A + B} 2$

$=$

$\ds \dfrac {\cos \frac {a - b} 2} {\cos \frac {a + b} 2} \cot \dfrac C 2$

$\text {(1)}: \quad$

$\ds \leadsto \ \ $

$\ds \tan \frac {A + B} 2 \cos \frac {a + b} 2$

$=$

$\ds \cos \frac {a - b} 2 \cot \frac C 2$

more manageable in this form

$\ds \tan \dfrac {A - B} 2$

$=$

$\ds \dfrac {\sin \frac {a - b} 2} {\sin \frac {a + b} 2} \cot \dfrac C 2$

$\text {(2)}: \quad$

$\ds \leadsto \ \ $

$\ds \tan \frac {A - B} 2 \sin \frac {a + b} 2$

$=$

$\ds \sin \frac {a - b} 2 \cot \frac C 2$

more manageable in this form

Hence we have:

$\ds \dfrac {\tan \frac {A + B} 2 \cos \frac {a + b} 2} {\tan \frac {A - B} 2 \sin \frac {a + b} 2}$

$=$

$\ds \dfrac {\cos \frac {a - b} 2 \cot \frac C 2} {\sin \frac {a - b} 2 \cot \frac C 2}$

dividing $(1)$ by $(2)$

$\ds \leadsto \ \ $

$\ds \dfrac {\tan \frac {A + B} 2} {\tan \frac {A - B} 2} \dfrac 1 {\tan \frac {a + b} 2}$

$=$

$\ds \dfrac 1 {\tan \frac {a - b} 2}$

simplifying

$\ds \leadsto \ \ $

$\ds \dfrac {\tan \frac {A + B} 2} {\tan \frac {A - B} 2}$

$=$

$\ds \dfrac {\tan \frac {a + b} 2} {\tan \frac {a - b} 2}$

simplifying

$\blacksquare$

1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 5$: Trigonometric Functions: $5.98$
2009: Murray R. Spiegel, Seymour Lipschutz and John Liu: Mathematical Handbook of Formulas and Tables (3rd ed.) ... (previous) ... (next): $\S 12$: Trigonometric Functions: $12.98.$