Theorem: Napier's Tangent Rule for Quadrantal Triangles

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Let $\triangle ABC$ be a quadrantal 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.
Let the side $c$ be a right angle.
Let the remaining parts of $\triangle ABC$ be arranged according to the '''exterior''' of this circle, where the symbol $\Box$ denotes a right angle.
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Let one of the parts of this circle be called a '''middle part'''.
Let the two neighboring parts of the '''middle part''' be called '''adjacent parts'''.
Then the sine of the '''middle part''' equals the product of the tangents of the '''adjacent parts'''.

Proof:
We are given that $c$ is a right angle.
Let the remaining parts of $\triangle ABC$ be arranged according to the '''exterior''' of the circle above, where the symbol $\Box$ denotes a right angle.
==== $\sin A$ ====
{{begin-eqn}}
{{eqn | l = \cos A \cos c
      | r = \sin A \cot B - \sin c \cot b
      | c = Four-Parts Formula on $b, A, c, B$
}}
{{eqn | ll= \leadsto
      | l = \cos A \times 0
      | r = \sin A \cot B - 1 \times \cot b
      | c = Cosine of Right Angle, Sine of Right Angle as $c = \Box$
}}
{{eqn | ll= \leadsto
      | l = \sin A \cot B
      | r = \cot b
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \sin A
      | r = \tan B \cot b
      | c = multiplying both sides by $\tan B = \dfrac 1 {\cot B}$
}}
{{eqn | ll= \leadsto
      | l = \sin A
      | r = \tan B \, \map \tan {\Box - b}
      | c = Tangent of Complement equals Cotangent
}}
{{end-eqn}}
{{qed|lemma}}
==== $\sin B$ ====
{{begin-eqn}}
{{eqn | l = \cos B \cos c
      | r = \sin B \cot A - \sin c \cot a
      | c = Four-Parts Formula on $A, c, B, a$
}}
{{eqn | ll= \leadsto
      | l = \cos B \times 0
      | r = \sin B \cot A - 1 \times \cot a
      | c = Cosine of Right Angle, Sine of Right Angle as $c = \Box$
}}
{{eqn | ll= \leadsto
      | l = \sin B \cot A
      | r = \cot a
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \sin B
      | r = \tan A \cot a
      | c = multiplying both sides by $\tan A = \dfrac 1 {\cot A}$
}}
{{eqn | ll= \leadsto
      | l = \sin B
      | r = \tan A \, \map \tan {\Box - a}
      | c = Tangent of Complement equals Cotangent
}}
{{end-eqn}}
{{qed|lemma}}
==== $\map \sin {\Box - a}$ ====
{{begin-eqn}}
{{eqn | l = \sin A \cos c
      | r = \cos C \sin B + \sin C \cos B \cos a
      | c = Analogue Formula for Spherical Law of Cosines:Corollary for side $a$
}}
{{eqn | ll= \leadsto
      | l = \sin A \times 0
      | r = \cos C \sin B + \sin C \cos B \cos a
      | c = Cosine of Right Angle as $c = \Box$
}}
{{eqn | ll= \leadsto
      | l = \sin C \cos B \cos a
      | r = -\cos C \sin B
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \cos a
      | r = -\cot C \tan B
      | c = dividing both sides by $\sin C \cos B$
}}
{{eqn | ll= \leadsto
      | l = \map \sin {\Box - a}
      | r = \map \tan {C - \Box} \tan b
      | c = Sine of Complement equals Cosine, Sine Function is Odd, Tangent of Complement equals Cotangent
}}
{{end-eqn}}
{{qed|lemma}}
==== $\map \sin {C - \Box}$ ====
{{begin-eqn}}
{{eqn | l = \cos c
      | r = \cos a \cos b + \sin a \sin b \cos C
      | c = Spherical Law of Cosines for side $c$
}}
{{eqn | ll= \leadsto
      | l = 0
      | r = \cos a \cos b + \sin a \sin b \cos C
      | c = Cosine of Right Angle as $c = \Box$
}}
{{eqn | ll= \leadsto
      | l = \sin a \sin b \cos C
      | r = -\cos a \cos b
      | c = 
}}
{{eqn | ll= \leadsto
      | l = -\cos C
      | r = \cot a \cot b
      | c = dividing both sides by $-\sin a \sin b$
}}
{{eqn | ll= \leadsto
      | l = \map \sin {C - \Box}
      | r = \map \tan {\Box - a} \, \map \tan {\Box - b} 
      | c = Sine of Complement equals Cosine, Sine Function is Odd, Tangent of Complement equals Cotangent
}}
{{end-eqn}}
{{qed|lemma}}
==== $\map \sin {\Box - b}$ ====
{{begin-eqn}}
{{eqn | l = \sin B \cos c
      | r = \cos C \sin A + \sin C \cos A \cos b
      | c = Analogue Formula for Spherical Law of Cosines:Corollary for side $b$
}}
{{eqn | ll= \leadsto
      | l = \sin B \times 0
      | r = \cos C \sin A + \sin C \cos A \cos b
      | c = Cosine of Right Angle as $c = \Box$
}}
{{eqn | ll= \leadsto
      | l = \sin C \cos A \cos b
      | r = -\cos C \sin A
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \cos b
      | r = -\cot C \tan A
      | c = dividing both sides by $\sin C \cos A$
}}
{{eqn | ll= \leadsto
      | l = \map \sin {\Box - b}
      | r = \map \tan {C - \Box} \tan a
      | c = Sine of Complement equals Cosine, Sine Function is Odd, Tangent of Complement equals Cotangent
}}
{{end-eqn}}
{{qed}}