Napier's Analogies

Let a Spherical Triangle have sides $a$, $b$, and $c$ with $A$, $B$, and $C$ the corresponding opposite angles. Then

$${\sin[{\textstyle{1\over 2}}(A-B)]\over\sin[{\textstyle{1\over 2}}(A+B)]} = {\tan[{\textstyle{1\over 2}}(a-b)]\over\tan({\textstyle{1\over 2}}c)}$$ (1)

$${\cos[{\textstyle{1\over 2}}(A-B)]\over\cos[{\textstyle{1\over 2}}(A+B)]} = {\tan[{\textstyle{1\over 2}}(a+b)]\over\tan({\textstyle{1\over 2}}c)}$$ (2)

$${\sin[{\textstyle{1\over 2}}(a-b)]\over\sin[{\textstyle{1\over 2}}(a+b)]} = {\tan[{\textstyle{1\over 2}}(A-B)]\over\cot({\textstyle{1\over 2}}C)}$$ (3)

$${\cos[{\textstyle{1\over 2}}(a-b)]\over\cos[{\textstyle{1\over 2}}(a+b)]} = {\tan[{\textstyle{1\over 2}}(A+B)]\over\cot({\textstyle{1\over 2}}C)}.$$ (4)

See also Spherical Trigonometry