avewright/proofwiki-math huggingface dataset:
Theorem: Cosine of 225 Degrees

:$\cos 225 \degrees = \cos \dfrac {5 \pi} 4 = -\dfrac {\sqrt 2} 2$

Proof:
{{begin-eqn}}
{{eqn | l = \cos 225 \degrees
      | r = \map \cos {360 \degrees - 135 \degrees}
      | c = 
}}
{{eqn | r = \cos 135 \degrees
      | c = Cosine of Conjugate Angle
}}
{{eqn | r = -\frac {\sqrt 2} 2
      | c = Cosine of $135 \degrees$
}}
{{end-eqn}}
{{qed}}
Theorem: Cosine of 240 Degrees

:$\cos 240 \degrees = \cos \dfrac {4 \pi} 3 = -\dfrac 1 2$

Proof:
{{begin-eqn}}
{{eqn | l = \cos 240 \degrees
      | r = \map \cos {360 \degrees - 120 \degrees}
      | c = 
}}
{{eqn | r = \cos 120 \degrees
      | c = Cosine of Conjugate Angle
}}
{{eqn | r = -\frac 1 2
      | c = Cosine of $120 \degrees$
}}
{{end-eqn}}
{{qed}}
Theorem: Cosine of 255 Degrees

:$\cos 255^\circ = \cos \dfrac {17 \pi} {12} = - \dfrac {\sqrt 6 - \sqrt 2} 4$

Proof:
{{begin-eqn}}
{{eqn | l = \cos 255^\circ
      | r = \cos \left({360^\circ - 105^\circ}\right)
      | c = 
}}
{{eqn | r = \cos 105^\circ
      | c = Cosine of Conjugate Angle
}}
{{eqn | r = - \frac {\sqrt 6 - \sqrt 2} 4
      | c = Cosine of 105 Degrees
}}
{{end-eqn}}
{{qed}}
Theorem: Cosine of Three Right Angles

:$\cos 270 \degrees = \cos \dfrac {3 \pi} 2 = 0$

Proof:
{{begin-eqn}}
{{eqn | l = \cos 270 \degrees
      | r = \map \cos {360 \degrees - 90 \degrees}
      | c = 
}}
{{eqn | r = \cos 90 \degrees
      | c = Cosine of Conjugate Angle
}}
{{eqn | r = 0
      | c = Cosine of Right Angle
}}
{{end-eqn}}
{{qed}}
Theorem: Cosine of 285 Degrees

:$\cos 285^\circ = \cos \dfrac {19 \pi} {12} = \dfrac {\sqrt 6 - \sqrt 2} 4$

Proof:
{{begin-eqn}}
{{eqn | l = \cos 285^\circ
      | r = \cos \left({360^\circ - 75^\circ}\right)
      | c = 
}}
{{eqn | r = \cos 75^\circ
      | c = Cosine of Conjugate Angle
}}
{{eqn | r = \frac {\sqrt 6 - \sqrt 2} 4
      | c = Cosine of 75 Degrees
}}
{{end-eqn}}
{{qed}}
Theorem: Cosine of 300 Degrees

:$\cos 300 \degrees = \cos \dfrac {5 \pi} 3 = \dfrac 1 2$

Proof:
{{begin-eqn}}
{{eqn | l = \cos 300 \degrees
      | r = \map \cos {360 \degrees - 60 \degrees}
      | c = 
}}
{{eqn | r = \cos 60 \degrees
      | c = Cosine of Conjugate Angle
}}
{{eqn | r = \frac 1 2
      | c = {{cos|60}}
}}
{{end-eqn}}
{{qed}}
Theorem: Cosine of 315 Degrees

:$\cos 315 \degrees = \cos \dfrac {7 \pi} 4 = \dfrac {\sqrt 2} 2$

Proof:
{{begin-eqn}}
{{eqn | l = \cos 315 \degrees
      | r = \map \cos {360 \degrees - 45 \degrees}
      | c = 
}}
{{eqn | r = \cos 45 \degrees
      | c = Cosine of Conjugate Angle
}}
{{eqn | r = \frac {\sqrt 2} 2
      | c = Cosine of $45 \degrees$
}}
{{end-eqn}}
{{qed}}
Theorem: Cosine of 330 Degrees

:$\cos 330 \degrees = \cos \dfrac {11 \pi} 6 = \dfrac {\sqrt 3} 2$

Proof:
{{begin-eqn}}
{{eqn | l = \cos 330 \degrees
      | r = \map \cos {360 \degrees - 30 \degrees}
      | c = 
}}
{{eqn | r = \cos 30 \degrees
      | c = Cosine of Conjugate Angle
}}
{{eqn | r = \frac {\sqrt 3} 2
      | c = Cosine of $30 \degrees$
}}
{{end-eqn}}
{{qed}}
Theorem: Cosine of 345 Degrees

:$\cos 345 \degrees = \cos \dfrac {23 \pi} {12} = \dfrac {\sqrt 6 + \sqrt 2} 4$

Proof:
{{begin-eqn}}
{{eqn | l = \cos 345 \degrees
      | r = \map \cos {360 \degrees - 15 \degrees}
      | c = 
}}
{{eqn | r = \cos 15 \degrees
      | c = Cosine of Conjugate Angle
}}
{{eqn | r = \frac {\sqrt 6 + \sqrt 2} 4
      | c = Cosine of $15 \degrees$
}}
{{end-eqn}}
{{qed}}
Theorem: Cosine of Full Angle

:$\cos 360 \degrees = \cos 2 \pi = 1$

Proof:
A direct implementation of Cosine of Multiple of Pi:
:$\forall n \in \Z: \cos n \pi = \paren {-1}^n$
In this case, $n = 2$ and so:
:$\cos 2 \pi = \paren {-1}^2 = 1$
{{qed}}
Theorem: Tangent of Zero

:$\tan 0 = 0$

Proof:
{{begin-eqn}}
{{eqn | l = \tan 0
      | r = \frac {\sin 0} {\cos 0}
      | c = Tangent is Sine divided by Cosine
}}
{{eqn | r = \frac 0 1
      | c = Sine of Zero is Zero and Cosine of Zero is One
}}
{{eqn | r = 0
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Tangent of 15 Degrees

:$\tan 15^\circ = \tan \dfrac {\pi} {12} = 2 - \sqrt 3$

Proof:
{{begin-eqn}}
{{eqn | l = \tan 15 \degrees
      | r = \tan \frac {30 \degrees} 2
      | c = 
}}
{{eqn | r = \frac {\sin 30 \degrees} {1 + \cos 30 \degrees}
      | c = {{Corollary|Half Angle Formula for Tangent|1}}
}}
{{eqn | r = \frac {\frac 1 2} {1 + \frac {\sqrt 3} 2}
      | c = {{sin|30}} and {{cos|30}}
}}
{{eqn | r = \frac {\frac 1 2} {\frac {2 + \sqrt 3} 2}
      | c = 
}}
{{eqn | r = \frac 1 {2 + \sqrt 3}
      | c = 
}}
{{eqn | r = \frac {2 - \sqrt 3} {\paren {2 + \sqrt 3} \paren {2 - \sqrt 3} }
      | c = multiplying top and bottom by $2 - \sqrt 3$
}}
{{eqn | r = \frac {2 - \sqrt 3} {4 - 3}
      | c = Difference of Two Squares
}}
{{eqn | r = 2 - \sqrt 3
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Tangent of 15 Degrees

:$\tan 15^\circ = \tan \dfrac {\pi} {12} = 2 - \sqrt 3$

Proof:
{{begin-eqn}}
{{eqn | l = \tan 15 \degrees
      | r = \tan \frac {30 \degrees} 2
      | c = 
}}
{{eqn | r = \frac {1 - \cos 30 \degrees} {\sin 30 \degrees}
      | c = {{Corollary|Half Angle Formula for Tangent|2}}
}}
{{eqn | r = \frac {1 - \frac {\sqrt 3} 2} {\frac 1 2} 
      | c = {{cos|30}} and {{sin|30}}
}}
{{eqn | r = 2 - \sqrt 3
      | c = multiplying top and bottom by $2$
}}
{{end-eqn}}
{{qed}}
Theorem: Tangent of 15 Degrees

:$\tan 15^\circ = \tan \dfrac {\pi} {12} = 2 - \sqrt 3$

Proof:
{{begin-eqn}}
{{eqn | l = \tan 15 \degrees
      | r = \frac {\sin 15 \degrees} {\cos 15 \degrees}
      | c = Tangent is Sine divided by Cosine
}}
{{eqn | r = \frac {\frac {\sqrt 6 - \sqrt 2} 4} {\frac {\sqrt 6 + \sqrt 2} 4}
      | c = {{sin|15}} and {{cos|15}}
}}
{{eqn | r = \frac {\sqrt 6 - \sqrt 2} {\sqrt 6 + \sqrt 2}
      | c = simplifying
}}
{{eqn | r = \frac {\paren {\sqrt 6 - \sqrt 2}^2} {\paren {\sqrt 6 + \sqrt 2} \paren {\sqrt 6 - \sqrt 2} }
      | c = multiplying top and bottom by $\sqrt 6 - \sqrt 2$ 
}}
{{eqn | r = \frac {6 - 2 \sqrt 6 \sqrt 2 + 2 } {6 - 2}
      | c = multiplying out, and Difference of Two Squares
}}
{{eqn | r = \frac {8 - 4 \sqrt 3} 4
      | c = simplifying
}}
{{eqn | r = 2 - \sqrt 3
      | c = dividing top and bottom by $4$
}}
{{end-eqn}}
{{qed}}
Theorem: Tangent of 30 Degrees

:$\tan 30 \degrees = \tan \dfrac \pi 6 = \dfrac {\sqrt 3} 3$

Proof:
{{begin-eqn}}
{{eqn | l = \tan 30 \degrees
      | r = \frac {\sin 30 \degrees} {\cos 30 \degrees}
      | c = Tangent is Sine divided by Cosine
}}
{{eqn | r = \frac {\frac 1 2} {\frac {\sqrt 3} 2}
      | c = Sine of $30 \degrees$ and Cosine of $30 \degrees$
}}
{{eqn | r = \frac 1 {\sqrt 3}
      | c = 
}}
{{eqn | r = \frac {\sqrt 3} 3
      | c = multiplying top and bottom by $\sqrt 3$
}}
{{end-eqn}}
{{qed}}
Theorem: Tangent of 45 Degrees

:$\tan 45 \degrees = \tan \dfrac \pi 4 = 1$

Proof:
{{begin-eqn}}
{{eqn | l = \tan 45 \degrees
      | r = \frac {\sin 45 \degrees} {\cos 45 \degrees}
      | c = Tangent is Sine divided by Cosine
}}
{{eqn | r = \frac {\frac {\sqrt 2} 2} {\frac {\sqrt 2} 2}
      | c = Sine of $45 \degrees$ and Cosine of $45 \degrees$
}}
{{eqn | r = 1
      | c = dividing top and bottom by $\sqrt 2 / 2$
}}
{{end-eqn}}
{{qed}}
Theorem: Tangent of 60 Degrees

:$\tan 60 \degrees = \tan \dfrac \pi 3 = \sqrt 3$

Proof:
{{begin-eqn}}
{{eqn | l = \tan 60 \degrees
      | r = \frac {\sin 60 \degrees} {\cos 60 \degrees}
      | c = Tangent is Sine divided by Cosine
}}
{{eqn | r = \frac {\frac {\sqrt 3} 2} {\frac 1 2}
      | c = Sine of $60 \degrees$ and Cosine of $60 \degrees$
}}
{{eqn | r = \sqrt 3
      | c = multiplying top and bottom by $2$
}}
{{end-eqn}}
{{qed}}
Theorem: Tangent of 75 Degrees

:$\tan 75 \degrees = \tan \dfrac {5 \pi} {12} = 2 + \sqrt 3$

Proof:
{{begin-eqn}}
{{eqn | l = \tan 75 \degrees
      | r = \frac {\sin 75 \degrees} {\cos 75 \degrees}
      | c = Tangent is Sine divided by Cosine
}}
{{eqn | r = \frac {\frac {\sqrt 6 + \sqrt 2} 4} {\frac {\sqrt 6 - \sqrt 2} 4}
      | c = {{sin|75}} and {{cos|75}}
}}
{{eqn | r = \frac {\sqrt 6 + \sqrt 2} {\sqrt 6 - \sqrt 2}
      | c = simplifying
}}
{{eqn | r = \frac {\paren {\sqrt 6 + \sqrt 2}^2} {\paren {\sqrt 6 - \sqrt 2} \paren {\sqrt 6 + \sqrt 2} }
      | c = multiplying top and bottom by $\sqrt 6 + \sqrt 2$ 
}}
{{eqn | r = \frac {6 + 2 \sqrt 6 \sqrt 2 + 2} {6 - 2}
      | c = multiplying out, and Difference of Two Squares
}}
{{eqn | r = \frac {8 + 4 \sqrt 3} 4
      | c = simplifying
}}
{{eqn | r = 2 + \sqrt 3
      | c = dividing top and bottom by $4$
}}
{{end-eqn}}
{{qed}}
Theorem: Tangent of Complement equals Cotangent

:$\map \tan {\dfrac \pi 2 - \theta} = \cot \theta$ for $\theta \ne n \pi$
where $\tan$ and $\cot$ are tangent and cotangent respectively.
That is, the cotangent of an angle is the tangent of its complement.
This relation is defined wherever $\sin \theta \ne 0$.

Proof:
{{begin-eqn}}
{{eqn | l = \map \tan {\frac \pi 2 - \theta}
      | r = \frac {\map \sin {\frac \pi 2 - \theta} } {\map \cos {\frac \pi 2 - \theta} }
      | c = Tangent is Sine divided by Cosine
}}
{{eqn | r = \frac {\cos \theta} {\sin \theta}
      | c = Sine and Cosine of Complementary Angles
}}
{{eqn | r = \cot \theta
      | c = Cotangent is Cosine divided by Sine
}}
{{end-eqn}}
The above is valid only where $\sin \theta \ne 0$, as otherwise $\dfrac {\cos \theta} {\sin \theta}$ is undefined.
From Sine of Multiple of Pi it follows that this happens when $\theta \ne n \pi$.
{{qed}}
Theorem: Cotangent of Complement equals Tangent

Let $\theta \ne \paren {2 n + 1} \dfrac \pi 2$
Then:
:$\map \cot {\dfrac \pi 2 - \theta} = \tan \theta$

Proof:
{{begin-eqn}}
{{eqn | l = \map \cot {\frac \pi 2 - \theta}
      | r = \frac {\map \cos {\frac \pi 2 - \theta} } {\map \sin {\frac \pi 2 - \theta} }
      | c = Cotangent is Cosine divided by Sine
}}
{{eqn | r = \frac {\sin \theta} {\cos \theta}
      | c = Sine and Cosine of Complementary Angles
}}
{{eqn | r = \tan \theta
      | c = Tangent is Sine divided by Cosine
}}
{{end-eqn}}
The above is valid only where $\cos \theta \ne 0$, as otherwise $\dfrac {\sin \theta} {\cos \theta}$ is undefined.
From Cosine of Half-Integer Multiple of Pi it follows that this happens when $\theta \ne \paren {2 n + 1} \dfrac \pi 2$.
{{qed}}
Theorem: Tangent of Angle plus Right Angle

:$\map \tan {x + \dfrac \pi 2} = -\cot x$

Proof:
{{begin-eqn}}
{{eqn | l = \map \tan {x + \frac \pi 2}
      | r = \frac {\map \sin {x + \frac \pi 2} } {\map \cos {x + \frac \pi 2} } 
      | c = Tangent is Sine divided by Cosine
}}
{{eqn | r = \frac {\cos x} {- \sin x}
      | c = Sine of Angle plus Right Angle and Cosine of Angle plus Right Angle
}}
{{eqn | r = -\cot x
      | c = Cotangent is Cosine divided by Sine
}}
{{end-eqn}}
{{qed}}
Theorem: Cotangent of Angle plus Right Angle

:$\map \cot {x + \dfrac \pi 2} = -\tan x$

Proof:
{{begin-eqn}}
{{eqn | l = \map \cot {x + \frac \pi 2}
      | r = \frac {\map \cos {x + \frac \pi 2} }  {\map \sin {x + \frac \pi 2} }
      | c = Cotangent is Cosine divided by Sine
}}
{{eqn | r = \frac {-\sin x} {\cos x}
      | c = Cosine of Angle plus Right Angle and Sine of Angle plus Right Angle
}}
{{eqn | r = -\tan x
      | c = Tangent is Sine divided by Cosine
}}
{{end-eqn}}
{{qed}}
Theorem: Tangent of Supplementary Angle

:$\map \tan {\pi - \theta} = -\tan \theta$
where $\tan$ denotes tangent.
That is, the tangent of an angle is the negative of its supplement.

Proof:
{{begin-eqn}}
{{eqn | l = \map \tan {\pi - \theta}
      | r = \frac {\map \sin {\pi - \theta} } {\map \cos {\pi - \theta} }
      | c = Tangent is Sine divided by Cosine
}}
{{eqn | r = \frac {\sin \theta} {-\cos \theta}
      | c = Sine of Supplementary Angle and Cosine of Supplementary Angle
}}
{{eqn | r = -\tan \theta
}}
{{end-eqn}}
{{qed}}
Theorem: Cotangent of Supplementary Angle

:$\map \cot {\pi - \theta} = -\cot \theta$
where $\cot$ denotes tangent.
That is, the cotangent of an angle is the negative of its supplement.

Proof:
{{begin-eqn}}
{{eqn | l = \map \cot {\pi - \theta}
      | r = \frac {\map \cos {\pi - \theta} } {\map \sin {\pi - \theta} }
      | c = Cotangent is Cosine divided by Sine
}}
{{eqn | r = \frac {-\cos \theta} {\sin \theta} 
      | c = Cosine of Supplementary Angle and Sine of Supplementary Angle
}}
{{eqn | r = -\cot \theta
}}
{{end-eqn}}
{{qed}}
Theorem: Tangent of Conjugate Angle

:$\map \tan {2 \pi - \theta} = -\tan \theta$
where $\tan$ denotes tangent.
That is, the tangent of an angle is the negative of its conjugate.

Proof:
{{begin-eqn}}
{{eqn | l = \map \tan {2 \pi - \theta}
      | r = \frac {\map \sin {2 \pi - \theta} } {\map \cos {2 \pi - \theta} }
      | c = Tangent is Sine divided by Cosine
}}
{{eqn | r = \frac {-\sin \theta} {\cos \theta}
      | c = Sine of Conjugate Angle and Cosine of Conjugate Angle
}}
{{eqn | r = -\tan \theta
      | c = Tangent is Sine divided by Cosine
}}
{{end-eqn}}
{{qed}}
Theorem: Cotangent of Conjugate Angle

:$\map \cot {2 \pi - \theta} = -\cot \theta$
where $\cot$ denotes cotangent.
That is, the cotangent of an angle is the negative of its conjugate.

Proof:
{{begin-eqn}}
{{eqn | l = \map \cot {2 \pi - \theta}
      | r = \frac {\map \cos {2 \pi - \theta} } {\map \sin {2 \pi - \theta} }
      | c = Cotangent is Cosine divided by Sine
}}
{{eqn | r = \frac {\cos \theta} {-\sin \theta}
      | c = Cosine of Conjugate Angle and Sine of Conjugate Angle
}}
{{eqn | r = -\cot \theta
      | c = Cotangent is Cosine divided by Sine
}}
{{end-eqn}}
{{qed}}
Theorem: Tangent of Right Angle

:$\tan 90 \degrees = \tan \dfrac \pi 2$ is undefined

Proof:
From Tangent is Sine divided by Cosine:
:$\tan \theta = \dfrac {\sin \theta} {\cos \theta}$
When $\cos \theta = 0$, $\dfrac {\sin \theta} {\cos \theta}$ can be defined only if $\sin \theta = 0$.
But there are no such $\theta$ such that both $\cos \theta = 0$ and $\sin \theta = 0$.
When $\theta = \dfrac \pi 2$, $\cos \theta = 0$.
Thus $\tan \theta$ is undefined at this value.
{{qed}}
Theorem: Tangent of 105 Degrees

:$\tan 105^\circ = \tan \dfrac {7 \pi} {12} = - \left({2 + \sqrt 3}\right)$

Proof:
{{begin-eqn}}
{{eqn | l = \tan 105^\circ
      | r = \tan \left({90^\circ + 15^\circ}\right)
      | c = 
}}
{{eqn | r = - \cot 15^\circ
      | c = Tangent of Angle plus Right Angle
}}
{{eqn | r = - \left({2 + \sqrt 3}\right)
      | c = Cotangent of 15 Degrees
}}
{{end-eqn}}
{{qed}}
Theorem: Tangent of 120 Degrees

:$\tan 120 \degrees = \tan \dfrac {2 \pi} 3 = -\sqrt 3$

Proof:
{{begin-eqn}}
{{eqn | l = \tan 120 \degrees
      | r = \map \tan {90 \degrees + 30 \degrees}
      | c = 
}}
{{eqn | r = -\cot 30 \degrees
      | c = Tangent of Angle plus Right Angle
}}
{{eqn | r = -\sqrt 3
      | c = Cotangent of $30 \degrees$
}}
{{end-eqn}}
{{qed}}
Theorem: Tangent of 135 Degrees

:$\tan 135^\circ = \tan \dfrac {3 \pi} 4 = - 1$

Proof:
{{begin-eqn}}
{{eqn | l = \tan 135^\circ
      | r = \tan \left({90^\circ + 45^\circ}\right)
      | c = 
}}
{{eqn | r = - \cot 45^\circ
      | c = Tangent of Angle plus Right Angle
}}
{{eqn | r = - 1
      | c = Cotangent of 45 Degrees
}}
{{end-eqn}}
{{qed}}
Theorem: Tangent of 150 Degrees

:$\tan 150 \degrees = \tan \dfrac {5 \pi} 6 = -\dfrac {\sqrt 3} 3$

Proof:
{{begin-eqn}}
{{eqn | l = \tan 150 \degrees
      | r = \map \tan {90 \degrees + 60 \degrees}
      | c = 
}}
{{eqn | r = -\cot 60 \degrees
      | c = Tangent of Angle plus Right Angle
}}
{{eqn | r = -\dfrac {\sqrt 3} 3
      | c = Cotangent of $60 \degrees$
}}
{{end-eqn}}
{{qed}}
Theorem: Tangent of 165 Degrees

:$\tan 165 \degrees = \tan \dfrac {11 \pi} {12} = -\paren {2 - \sqrt 3}$

Proof:
{{begin-eqn}}
{{eqn | l = \tan 165 \degrees
      | r = \map \tan {90 \degrees + 75 \degrees}
      | c = 
}}
{{eqn | r = -\cot 75 \degrees
      | c = Tangent of Angle plus Right Angle
}}
{{eqn | r = -\paren {2 - \sqrt 3}
      | c = Cotangent of $75 \degrees$
}}
{{end-eqn}}
{{qed}}
Theorem: Tangent of Straight Angle

:$\tan 180 \degrees = \tan \pi = 0$

Proof:
{{begin-eqn}}
{{eqn | l = \tan 180 \degrees
      | r = \frac {\sin 180 \degrees} {\cos 180 \degrees}
      | c = Tangent is Sine divided by Cosine
}}
{{eqn | r = \frac 0 {-1}
      | c = Sine of Straight Angle and Cosine of Straight Angle
}}
{{eqn | r = 0
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Tangent of 195 Degrees

:$\tan 195 \degrees = \tan \dfrac {13 \pi} {12} = 2 - \sqrt 3$

Proof:
{{begin-eqn}}
{{eqn | l = \tan 195 \degrees
      | r = \map \tan {360 \degrees - 165 \degrees}
      | c = 
}}
{{eqn | r = -\tan 165 \degrees
      | c = Tangent of Conjugate Angle
}}
{{eqn | r = 2 - \sqrt 3
      | c = Tangent of $165 \degrees$
}}
{{end-eqn}}
{{qed}}
Theorem: Tangent of 210 Degrees

:$\tan 210 \degrees = \tan \dfrac {7 \pi} 6 = \dfrac {\sqrt 3} 3$

Proof:
{{begin-eqn}}
{{eqn | l = \tan 210 \degrees
      | r = \map \tan {360 \degrees - 150 \degrees}
      | c = 
}}
{{eqn | r = -\tan 150 \degrees
      | c = Tangent of Conjugate Angle
}}
{{eqn | r = \frac {\sqrt 3} 3
      | c = Tangent of $150 \degrees$
}}
{{end-eqn}}
{{qed}}
Theorem: Tangent of 225 Degrees

:$\tan 225 \degrees = \tan \dfrac {5 \pi} 4 = 1$

Proof:
{{begin-eqn}}
{{eqn | l = \tan 225 \degrees
      | r = \map \tan {360 \degrees - 135 \degrees}
      | c = 
}}
{{eqn | r = -\tan 135 \degrees
      | c = Tangent of Conjugate Angle
}}
{{eqn | r = 1
      | c = {{tan|135}}
}}
{{end-eqn}}
{{qed}}
Theorem: Tangent of 240 Degrees

:$\tan 240 \degrees = \tan \dfrac {4 \pi} 3 = \sqrt 3$

Proof:
{{begin-eqn}}
{{eqn | l = \tan 240 \degrees
      | r = \map \tan {360 \degrees - 120 \degrees}
      | c = 
}}
{{eqn | r = -\tan 120 \degrees
      | c = Tangent of Conjugate Angle
}}
{{eqn | r = \sqrt 3
      | c = Tangent of $120 \degrees$
}}
{{end-eqn}}
{{qed}}
Theorem: Tangent of 255 Degrees

:$\tan 255 \degrees = \tan \dfrac {17 \pi} {12} = 2 + \sqrt 3$

Proof:
{{begin-eqn}}
{{eqn | l = \tan 255 \degrees
      | r = \map \tan {360 \degrees - 105 \degrees}
      | c = 
}}
{{eqn | r = -\tan 105 \degrees
      | c = Tangent of Conjugate Angle
}}
{{eqn | r = 2 + \sqrt 3
      | c = Tangent of $105 \degrees$
}}
{{end-eqn}}
{{qed}}
Theorem: Tangent of Three Right Angles

:$\tan 270 \degrees = \tan \dfrac {3 \pi} 2$ is undefined

Proof:
We have:
{{begin-eqn}}
{{eqn | l = \tan 270 \degrees
      | r = \map \tan {360 \degrees - 90 \degrees}
      | c = 
}}
{{eqn | r = -\tan 90 \degrees
      | c = Tangent of Conjugate Angle
}}
{{end-eqn}}
But from Tangent of Right Angle, $\tan 90 \degrees$ is undefined.
Hence so is $\tan 270 \degrees$.
{{qed}}
Theorem: Tangent of 285 Degrees

:$\tan 285 \degrees = \tan \dfrac {19 \pi} {12} = -\paren {2 + \sqrt 3}$

Proof:
{{begin-eqn}}
{{eqn | l = \tan 285 \degrees
      | r = \map \tan {360 \degrees - 75 \degrees}
      | c = 
}}
{{eqn | r = -\tan 75 \degrees
      | c = Tangent of Conjugate Angle
}}
{{eqn | r = -\paren {2 + \sqrt 3}
      | c = Tangent of $75 \degrees$
}}
{{end-eqn}}
{{qed}}
Theorem: Tangent of 300 Degrees

:$\tan 300 \degrees = \tan \dfrac {5 \pi} 3 = -\sqrt 3$

Proof:
{{begin-eqn}}
{{eqn | l = \tan 300 \degrees
      | r = \map \tan {360 \degrees - 60 \degrees}
      | c = 
}}
{{eqn | r = -\tan 60 \degrees
      | c = Tangent of Conjugate Angle
}}
{{eqn | r = -\sqrt 3
      | c = Tangent of $60 \degrees$
}}
{{end-eqn}}
{{qed}}
Theorem: Tangent of 315 Degrees

:$\tan 315 \degrees = \tan \dfrac {7 \pi} 4 = -1$

Proof:
{{begin-eqn}}
{{eqn | l = \tan 315 \degrees
      | r = \map \tan {360 \degrees - 45 \degrees}
      | c = 
}}
{{eqn | r = -\tan 45 \degrees
      | c = Tangent of Conjugate Angle
}}
{{eqn | r = -1
      | c = Tangent of $45 \degrees$
}}
{{end-eqn}}
{{qed}}
Theorem: Tangent of 330 Degrees

:$\tan 330^\circ = \tan \dfrac {11 \pi} 6 = -\dfrac {\sqrt 3} 3$

Proof:
{{begin-eqn}}
{{eqn | l = \tan 330^\circ
      | r = \tan \left({360^\circ - 30^\circ}\right)
      | c = 
}}
{{eqn | r = -\tan 30^\circ
      | c = Tangent of Conjugate Angle
}}
{{eqn | r = -\frac {\sqrt 3} 3
      | c = Tangent of 30 Degrees
}}
{{end-eqn}}
{{qed}}
Theorem: Tangent of 345 Degrees

:$\tan 345^\circ = \tan \dfrac {23 \pi} {12} = -\left({2 - \sqrt 3}\right)$

Proof:
{{begin-eqn}}
{{eqn | l = \tan 345^\circ
      | r = \tan \left({360^\circ - 15^\circ}\right)
      | c = 
}}
{{eqn | r = -\tan 15^\circ
      | c = Tangent of Conjugate Angle
}}
{{eqn | r = -\left({2 - \sqrt 3}\right)
      | c = Tangent of 15 Degrees
}}
{{end-eqn}}
{{qed}}
Theorem: Tangent of Full Angle

:$\tan 360^\circ = \tan 2 \pi = 0$

Proof:
{{begin-eqn}}
{{eqn | l = \tan 360^\circ
      | r = \tan \left({360^\circ - 0^\circ}\right)
      | c = 
}}
{{eqn | r = -\tan 0
      | c = Tangent of Conjugate Angle
}}
{{eqn | r = 0
      | c = Tangent of Zero
}}
{{end-eqn}}
{{qed}}
Theorem: Cotangent of Zero

:$\cot 0$ is undefined

Proof:
From Cotangent is Cosine divided by Sine:
:$\cot \theta = \dfrac {\cos \theta} {\sin \theta}$
When $\sin \theta = 0$, $\dfrac {\cos \theta} {\sin \theta}$ can be defined only if $\cos \theta = 0$.
But there are no such $\theta$ such that both $\cos \theta = 0$ and $\sin \theta = 0$.
When $\theta = 0$, $\sin \theta = 0$.
Thus $\cot \theta$ is undefined at this value.
{{qed}}
Theorem: Cotangent of 30 Degrees

:$\cot 30 \degrees = \cot \dfrac \pi 6 = \sqrt 3$

Proof:
{{begin-eqn}}
{{eqn | l = \cot 30 \degrees
      | r = \frac {\cos 30 \degrees} {\sin 30 \degrees}
      | c = Cotangent is Cosine divided by Sine
}}
{{eqn | r = \frac {\frac {\sqrt 3} 2} {\frac 1 2} 
      | c = {{cos|30}} and {{sin|30}}
}}
{{eqn | r = \sqrt 3
      | c = multiplying top and bottom by $2$
}}
{{end-eqn}}
{{qed}}
Theorem: Cotangent of 15 Degrees

:$\cot 15 \degrees = \cot \dfrac {\pi} {12} = 2 + \sqrt 3$

Proof:
{{begin-eqn}}
{{eqn | l = \cot 15^\circ
      | r = \frac {\cos 15 \degrees} {\sin 15 \degrees}
      | c = Cotangent is Cosine divided by Sine
}}
{{eqn | r = \frac {\frac {\sqrt 6 + \sqrt 2} 4} {\frac {\sqrt 6 - \sqrt 2} 4} 
      | c = {{cos|15}} and {{sin|15}}
}}
{{eqn | r = \frac {\sqrt 6 + \sqrt 2} {\sqrt 6 - \sqrt 2}
      | c = simplifying
}}
{{eqn | r = \frac {\paren {\sqrt 6 + \sqrt 2}^2} {\paren {\sqrt 6 - \sqrt 2} \paren {\sqrt 6 + \sqrt 2} }
      | c = multiplying top and bottom by $\sqrt 6 + \sqrt 2$ 
}}
{{eqn | r = \frac {6 + 2 \sqrt 6 \sqrt 2 + 2 } {6 - 2}
      | c = multiplying out, and Difference of Two Squares
}}
{{eqn | r = \frac {8 + 4 \sqrt 3} 4
      | c = simplifying
}}
{{eqn | r = 2 + \sqrt 3
      | c = dividing top and bottom by $4$
}}
{{end-eqn}}
{{qed}}
Theorem: Cotangent of 45 Degrees

:$\cot 45 \degrees = \cot \dfrac \pi 4 = 1$

Proof:
{{begin-eqn}}
{{eqn | l = \cot 45 \degrees
      | r = \frac {\cos 45 \degrees} {\sin 45 \degrees}
      | c = Cotangent is Cosine divided by Sine
}}
{{eqn | r = \frac {\frac {\sqrt 2} 2} {\frac {\sqrt 2} 2} 
      | c = {{cos|45}} and {{sin|45}}
}}
{{eqn | r = 1
      | c = dividing top and bottom by $\dfrac {\sqrt 2} 2$
}}
{{end-eqn}}
{{qed}}
Theorem: Cotangent of 60 Degrees

:$\cot 60 \degrees = \cot \dfrac \pi 3 = \dfrac {\sqrt 3} 3$

Proof:
{{begin-eqn}}
{{eqn | l = \cot 60 \degrees
      | r = \frac {\cos 60 \degrees} {\sin 60 \degrees}
      | c = Cotangent is Cosine divided by Sine
}}
{{eqn | r = \frac {\frac 1 2} {\frac {\sqrt 3} 2} 
      | c = {{cos|60}} and {{sin|60}}
}}
{{eqn | r = \frac {\sqrt 3} 3
      | c = multiplying top and bottom by $2 \sqrt 3$
}}
{{end-eqn}}
{{qed}}
Theorem: Cotangent of 75 Degrees

:$\cot 75 \degrees = \cot \dfrac {5 \pi} {12} = 2 - \sqrt 3$

Proof:
{{begin-eqn}}
{{eqn | l = \cot 75 \degrees
      | r = \frac {\cos 75 \degrees} {\sin 75 \degrees}
      | c = Cotangent is Cosine divided by Sine
}}
{{eqn | r = \frac {\frac {\sqrt 6 - \sqrt 2} 4} {\frac {\sqrt 6 + \sqrt 2} 4}
      | c = {{cos|75}} and {{sin|75}}
}}
{{eqn | r = \frac {\sqrt 6 - \sqrt 2} {\sqrt 6 + \sqrt 2}
      | c = simplifying
}}
{{eqn | r = \frac {\paren {\sqrt 6 - \sqrt 2}^2} {\paren {\sqrt 6 + \sqrt 2} \paren {\sqrt 6 - \sqrt 2} }
      | c = multiplying top and bottom by $\sqrt 6 - \sqrt 2$ 
}}
{{eqn | r = \frac {6 - 2 \sqrt 6 \sqrt 2 + 2} {6 - 2}
      | c = multiplying out, and Difference of Two Squares
}}
{{eqn | r = \frac {8 - 4 \sqrt 3} 4
      | c = simplifying
}}
{{eqn | r = 2 - \sqrt 3
      | c = dividing top and bottom by $4$
}}
{{end-eqn}}
{{qed}}
Theorem: Cotangent of Right Angle

:$\cot 90 \degrees = \cot \dfrac \pi 2 = 0$

Proof:
{{begin-eqn}}
{{eqn | l = \cot 90 \degrees
      | r = \frac {\cos 90 \degrees} {\sin 90 \degrees}
      | c = Cotangent is Cosine divided by Sine
}}
{{eqn | r = \frac 0 1
      | c = Cosine of Right Angle and Sine of Right Angle
}}
{{eqn | r = 0
      | c =
}}
{{end-eqn}}
{{qed}}