Spherical Law of Tangents

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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:

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


Proof

\(\ds \tan \dfrac {A + B} 2\) \(=\) \(\ds \dfrac {\cos \frac {a - b} 2} {\cos \frac {a + b} 2} \cot \dfrac C 2\) Napier's Analogies
\(\text {(1)}: \quad\) \(\ds \leadsto \ \ \) \(\ds \tan \frac {A + B} 2 \cos \frac {a + b} 2\) \(=\) \(\ds \cos \frac {a - b} 2 \cot \frac C 2\) more manageable in this form


\(\ds \tan \dfrac {A - B} 2\) \(=\) \(\ds \dfrac {\sin \frac {a - b} 2} {\sin \frac {a + b} 2} \cot \dfrac C 2\) Napier's Analogies
\(\text {(2)}: \quad\) \(\ds \leadsto \ \ \) \(\ds \tan \frac {A - B} 2 \sin \frac {a + b} 2\) \(=\) \(\ds \sin \frac {a - b} 2 \cot \frac C 2\) 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

$\blacksquare$


Sources

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