# 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:

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

## 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

$\blacksquare$