nyuuzyou/wikis huggingface dataset:
Scientific Notation/Examples/Multiplication/Arbitrary Example

== Example of Multiplication using Scientific Notation ==
:$\paren {4 \, 000 \, 000} \paren {0 \cdotp 000 \, 000 \, 000 \, 2} = 0 \cdotp 000 \, 8$
== Proof ==
== Sources ==
* : Chapter $1$: Scientific Notation: Example 2.
Category:Examples of Scientific Notation
Scientific Notation/Examples/Speed of Light

== Example of Scientific Notation ==
The speed of light is defined as:
:$c = 299 \, 792 \, 458 \text { m s}^{-1}$
In scientific notation this can be expressed as:
:$c = 2 \cdotp 99792 \, 458 \times 10^8 \text { m s}^{-1}$
== Sources ==
* : $2^{86243} - 1$
Category:Examples of Scientific Notation
Scope (Logic)/Examples/Arbitrary Example 2

== Example of Scope in the context of Logic ==
Consider the statement:
:$\paren {p \land \paren {q \lor r} } \implies \paren {s \iff \neg \, t}$
We have:
:$(1): \quad$ The scope of $\land$ is $p$ and $\paren {q \lor r}$.
:$(2): \quad$ The scope of $\lor$ is $q$ and $r$.
:$(3): \quad$ The scope of $\implies$ is $\paren {p \land \paren {q \lor r} }$ and $\paren {s \iff \neg \, t}$.
:$(4): \quad$ The scope of $\iff$ is $s$ and $\neg \, t$.
:$(5): \quad$ The scope of $\neg$ is $t$.
Category:Examples of Scope
Scope (Logic)/Examples/Arbitrary Example 3

== Example of Scope in the context of Logic ==
Consider the statement:
:$\exists x: \paren {x < y} \lor y = 0$
We have:
:$(1): \quad$ The scope of $\exists$ is $x$.
:$(2): \quad$ The scope of $\exists x$ is $\exists x: \paren {x < y}$.
:$(3): \quad$ The scope of $=$ is $y$ and $0$.
:$(4): \quad$ The scope of $\lor$ is $\exists x: \paren {x < y}$ and $y = 0$.
 
Scope (Logic)/Connective	Scope (Logic)/Connective 	Scope 1 	Scope 2
$ = $ 	$ y=0 $ 	$ y $ 	$ 0 $
$ \lor $ 	$\exists x: \paren { x < y } \lor y=0 $	$\exists x: \paren { x < y } $	$ y=0 $
Quantifier 	Scope (Logic)/Quantifier 	Scope 	
$\exists x$ 	$\exists x: \paren{ x < y } $ 	$ x < y $ 	
== Sources ==
* : $\S 1.2$ Logic and Notation
Category:Examples of Scope
Scott Topological Lattice is T0 Space

== Theorem ==
Let $T = \left({S, \preceq, \tau}\right)$ be a complete topological lattice with Scott topology.
Then $T$ is a $T_0$ space.
== Proof ==
Let $x, y \in S$ such that
:$x \ne y$
By Closure of Singleton is Lower Closure of Element in Scott Topological Lattice:
:$\left\{ {x}\right\}^- = x^\preceq$ and $\left\{ {y}\right\}^- = y^\preceq$
Thus by Lower Closures are Equal implies Elements are Equal:
:$\left\{ {x}\right\}^- \ne \left\{ {y}\right\}^-$
Hence by Characterization of T0 Space by Distinct Closures of Singletons:
:$T$ is $T_0$ space.
== Sources ==
* 
* 
Category:Topological Order Theory
Category:T0 Spaces
Scott Topology equals to Scott Sigma

== Theorem ==
Let $\struct {T, \preceq, \tau}$ be a up-complete topological lattice with Scott topology.
Then $\tau = \map \sigma {T, \preceq}$
where $\map \sigma L$ denotes the Scott sigma of $L$.
== Proof ==
This follows by Open iff Upper and with Property (S) in Scott Topological Lattice and definition Scott sigma.
== Sources ==
* 
Category:Topological Order Theory
Secant Function in terms of Gaussian Hypergeometric Function

== Theorem ==
:$\sec x = \map F {\dfrac 1 2, 1; 1; \sin^2 x}$
where:
:$x$ is a real number with $x \ne \dfrac {\paren {2 k + 1} \pi} 2$ with $k \in \Z$
:$\sec$ denotes the secant function
:$\map F {a, b; c; x}$ is the Gaussian hypergeometric function of $x$: $\ds \sum_{k \mathop = 0}^\infty \dfrac { a^{\overline k} b^{\overline k} } { c^{\overline k} } \dfrac {x^k} {k!}$
== Proof ==
== Also see ==
* Arcsine Function in terms of Gaussian Hypergeometric Function
* Arctangent Function in terms of Gaussian Hypergeometric Function
* Cosecant Function in terms of Gaussian Hypergeometric Function
== Sources ==
* : $\S 31$: Hypergeometric Functions: Special Cases: $31.7$
Category:Secant Function
Category:Gaussian Hypergeometric Function
Category:Hypergeometric Functions
Secant Function is Even

== Theorem ==
Let $x \in \R$ be a real number.
Let $\sec x$ be the secant of $x$.
Then, whenever $\sec x$ is defined:
:$\map \sec {-x} = \sec x$
That is, the secant function is even.
== Proof ==
== Also see ==
* Sine Function is Odd
* Cosine Function is Even
* Tangent Function is Odd
* Cotangent Function is Odd
* Cosecant Function is Odd
== Sources ==
* : $\S 5$: Trigonometric Functions: $5.32$
* : $\S 5$: Trigonometric Functions: Functions of Angles in All Quadrants in terms of those in Quadrant I
Category:Secant Function
Category:Examples of Even Functions
Secant Method can be derived from Newton-Raphson Method

== Theorem ==
Let $f: \R \to \R$ be a real function which has a root which is to be found.
The secant method can be derived from the Newton-Raphson method.
== Proof ==
We note that:
:$\map {f'} {x_n} \approx \dfrac {\map f x - \map f {x_{n - 1} } } {x_n - x_{n - 1} }$
The result follows by substituting for $\map {f'} {x_n}$ in the definition of the Newton-Raphson method.
== Sources ==
* 
* 
Category:Secant Method
Category:Newton-Raphson Method
Secant Secant Theorem

== Theorem ==
Let $C$ be a point external to a circle $ABED$.
Let $CA$ and $CB$ be straight lines which cut the circle at $D$ and $E$ respectively.
Then:
:$CA \cdot CD = CB \cdot CE$
== Proof ==
:
Draw $CF$ tangent to the circle.
From the Tangent Secant Theorem we have that:
:$CF^2 = CA \cdot CD$
:$CF^2 = CB \cdot CE$
from which the result is obvious and immediate.
== Also known as ==
== Also see ==
* Tangent Secant Theorem
* Intersecting Chords Theorem, of which this result is a generalization, where the point of intersection of the two lines is outside the circle.
* Power of a Point Theorem: a generalization of both.
Category:Secant Secant Theorem
Category:Circles
Category:Named Theorems
Secant and Cosecant are Cofunctions

== Theorem ==
The secant and cosecant are cofunctions:
== Proof ==
=== Proof of $(1)$ ===
From Secant is Reciprocal of Cosine:
:$\sec x = \dfrac 1 {\cos x}$
Hence in order for $\sec x$ to be defined it is necessary for $\cos x \ne 0$.
Then we have:
:Secant of Complement equals Cosecant
=== Proof of $(2)$ ===
From Cosecant is Reciprocal of Sine:
:$\csc x = \dfrac 1 {\sin x}$
Hence in order for $\csc x$ to be defined it is necessary for $\sin x \ne 0$.
Then we have:
:Cosecant of Complement equals Secant
Hence the result by definition of cofunction.
== Sources ==
* 
* 
Category:Cofunctions
Secant in terms of Hyperbolic Secant

== Theorem ==
Let $z \in \C$ be a complex number.
Then:
:$\sec z = \map \sech {i z}$
where:
:$\sec$ denotes the secant function
:$\sech$ denotes the hyperbolic secant
:$i$ is the imaginary unit: $i^2 = -1$.
== Proof ==
 
== Also see ==
* Sine in terms of Hyperbolic Sine
* Cosine in terms of Hyperbolic Cosine
* Tangent in terms of Hyperbolic Tangent
* Cotangent in terms of Hyperbolic Cotangent
* Cosecant in terms of Hyperbolic Cosecant
== Sources ==
* : $\S 8$: Hyperbolic Functions: $8.84$: Relationship between Hyperbolic and Trigonometric Functions
Category:Secant Function
Category:Hyperbolic Secant Function
Secant in terms of Tangent

== Theorem ==
Let $x$ be a real number such that $\cos x \ne 0$.
Then:
where $\sec$ denotes the real secant function and $\tan$ denotes the real tangent function.
== Proof ==
Also, from Sign of Secant:
:If there exists integer $n$ such that $\paren {2 n - \dfrac 1 2} \pi < x < \paren {2 n + \dfrac 1 2} \pi$, then $\sec x > 0$.
:If there exists integer $n$ such that $\paren {2 n + \dfrac 1 2} \pi < x < \paren {2 n + \dfrac 3 2} \pi$, then $\sec x < 0$.
When $\cos x = 0$, $\sec x$ and $\tan x$ is undefined.
== Also see ==
* Trigonometric Functions in terms of each other
Category:Secant Function
Category:Tangent Function
Secant is Reciprocal of Cosine

== Theorem ==
Let $\theta$ be an angle such that $\cos \theta \ne 0$.
Then:
:$\sec \theta = \dfrac 1 {\cos \theta}$
where $\sec$ and $\cos$ mean secant and cosine respectively.
== Proof ==
Let a point $P = \tuple {x, y}$ be placed in a cartesian plane with origin $O$ such that $OP$ forms an angle $\theta$ with the $x$-axis.
Then:
When $\cos \theta = 0$, $\dfrac 1 {\cos \theta}$ is not defined.
== Sources ==
* : $\text V$. Trigonometry: Definitions of the ratios
* : $\S 5$: Trigonometric Functions: $5.17$
* 
* 
* : Appendix $12$: Trigonometric formulae
* : Appendix $14$: Trigonometric formulae
Category:Cosine Function
Category:Secant Function
Secant of 105 Degrees

== Theorem ==
:$\sec 105 \degrees = \sec \dfrac {7 \pi} {12} = -\paren {\sqrt 6 + \sqrt 2}$
where $\sec$ denotes secant.
== Proof ==
== Sources ==
* : $\S 5$: Trigonometric Functions: Exact Values for Trigonometric Functions of Various Angles
Category:Secant Function
Secant of 135 Degrees

== Theorem ==
:$\sec 135 \degrees = \sec \dfrac {3 \pi} 4 = -\sqrt 2$
where $\sec$ denotes secant.
== Proof ==
== Sources ==
* : $\S 5$: Trigonometric Functions: Exact Values for Trigonometric Functions of Various Angles
Category:Secant Function
Secant of 150 Degrees

== Theorem ==
:$\sec 150 \degrees = \sec \dfrac {5 \pi} 6 = -\dfrac {2 \sqrt 3} 3$
where $\sec$ denotes secant.
== Proof ==
== Sources ==
* : $\S 5$: Trigonometric Functions: Exact Values for Trigonometric Functions of Various Angles
Category:Secant Function
Secant of 15 Degrees

== Theorem ==
:$\sec 15 \degrees = \sec \dfrac \pi {12} = \sqrt 6 - \sqrt 2$
where $\sec$ denotes secant.
== Proof ==
== Sources ==
* : $\S 5$: Trigonometric Functions: Exact Values for Trigonometric Functions of Various Angles
Category:Secant Function
Secant of 165 Degrees

== Theorem ==
:$\sec 165 \degrees = \sec \dfrac {11 \pi} {12} = -\paren {\sqrt 6 - \sqrt 2}$
where $\sec$ denotes secant.
== Proof ==
== Sources ==
* : $\S 5$: Trigonometric Functions: Exact Values for Trigonometric Functions of Various Angles
Category:Secant Function
Secant of 195 Degrees

== Theorem ==
:$\sec 195 \degrees = \sec \dfrac {13 \pi} {12} = -\paren {\sqrt 6 - \sqrt 2}$
where $\sec$ denotes secant.
== Proof ==
== Sources ==
* : $\S 5$: Trigonometric Functions: Exact Values for Trigonometric Functions of Various Angles
Category:Secant Function
Secant of 210 Degrees

== Theorem ==
:$\sec 210 \degrees = \sec \dfrac {7 \pi} 6 = -2 \dfrac {\sqrt 3} 3$
where $\sec$ denotes secant.
== Proof ==
== Sources ==
* : $\S 5$: Trigonometric Functions: Exact Values for Trigonometric Functions of Various Angles
Category:Secant Function
Secant of 225 Degrees

== Theorem ==
:$\sec 225 \degrees = \sec \dfrac {5 \pi} 4 = -\sqrt 2$
where $\sec$ denotes secant.
== Proof ==
== Sources ==
* : $\S 5$: Trigonometric Functions: Exact Values for Trigonometric Functions of Various Angles
Category:Secant Function
Secant of 255 Degrees

== Theorem ==
:$\sec 255 \degrees = \sec \dfrac {17 \pi} {12} = -\paren {\sqrt 6 + \sqrt 2}$
where $\sec$ denotes secant.
== Proof ==
== Sources ==
* : $\S 5$: Trigonometric Functions: Exact Values for Trigonometric Functions of Various Angles
Category:Secant Function
Secant of 285 Degrees

== Theorem ==
:$\sec 285 \degrees = \sec \dfrac {19 \pi} {12} = \sqrt 6 + \sqrt 2$
where $\sec$ denotes secant.
== Proof ==
== Sources ==
* : $\S 5$: Trigonometric Functions: Exact Values for Trigonometric Functions of Various Angles
Category:Secant Function
Secant of 30 Degrees

== Theorem ==
:$\sec 30 \degrees = \sec \dfrac \pi 6 = \dfrac {2 \sqrt 3} 3$
where $\sec$ denotes secant.
== Proof ==
== Sources ==
* : $\S 5$: Trigonometric Functions: Exact Values for Trigonometric Functions of Various Angles
Category:Secant Function
Secant of 315 Degrees

== Theorem ==
:$\sec 315 \degrees = \sec \dfrac {7 \pi} 4 = \sqrt 2$
where $\sec$ denotes secant.
== Proof ==
== Sources ==
* : $\S 5$: Trigonometric Functions: Exact Values for Trigonometric Functions of Various Angles
Category:Secant Function
Secant of 330 Degrees

== Theorem ==
:$\sec 330 \degrees = \sec \dfrac {11 \pi} 6 = 2 \dfrac {\sqrt 3} 3$
where $\sec$ denotes secant.
== Proof ==
== Sources ==
* : $\S 5$: Trigonometric Functions: Exact Values for Trigonometric Functions of Various Angles
Category:Secant Function
Secant of 345 Degrees

== Theorem ==
:$\sec 345 \degrees = \sec \dfrac {23 \pi} {12} = \sqrt 6 - \sqrt 2$
where $\sec$ denotes secant.
== Proof ==
== Sources ==
* : $\S 5$: Trigonometric Functions: Exact Values for Trigonometric Functions of Various Angles
Category:Secant Function
Secant of 75 Degrees

== Theorem ==
:$\sec 75 \degrees = \sec \dfrac {5 \pi} {12} = \sqrt 6 + \sqrt 2$
where $\sec$ denotes secant.
== Proof ==
== Sources ==
* : $\S 5$: Trigonometric Functions: Exact Values for Trigonometric Functions of Various Angles
Category:Secant Function
Secant of Angle in Cartesian Plane

== Theorem ==
Let $P = \tuple {x, y}$ be a point in the cartesian plane whose origin is at $O$.
Let $\theta$ be the angle between the $x$-axis and the line $OP$.
Let $r$ be the length of $OP$.
Then:
:$\sec \theta = \dfrac r x$
where $\sec$ denotes the secant of $\theta$.
== Proof ==
== Sources ==
* : $\S 5$: Trigonometric Functions: $5.11$
Category:Secant Function
Secant of Angle plus Full Angle

== Theorem ==
: $\sec \left({x + 2 \pi}\right) = \sec x$
== Proof ==
== Also see ==
* Sine of Angle plus Full Angle
* Cosine of Angle plus Full Angle
* Tangent of Angle plus Full Angle
* Cotangent of Angle plus Full Angle
* Cosecant of Angle plus Full Angle
== Sources ==
* : $\S 5$: Trigonometric Functions: Functions of Angles in All Quadrants in terms of those in Quadrant I
Category:Secant Function
Secant of Angle plus Right Angle

== Theorem ==
: $\sec \left({x + \dfrac \pi 2}\right) = -\csc x$
== Proof ==
== Also see ==
* Sine of Angle plus Right Angle
* Cosine of Angle plus Right Angle
* Tangent of Angle plus Right Angle
* Cotangent of Angle plus Right Angle
* Cosecant of Angle plus Right Angle
== Sources ==
* : $\S 5$: Trigonometric Functions: Functions of Angles in All Quadrants in terms of those in Quadrant I
Category:Secant Function
Category:Reduction Formulae (Trigonometry)
Secant of Angle plus Straight Angle

== Theorem ==
:$\map \sec {x + \pi} = -\sec x$
== Proof ==
== Also see ==
* Sine of Angle plus Straight Angle
* Cosine of Angle plus Straight Angle
* Tangent of Angle plus Straight Angle
* Cotangent of Angle plus Straight Angle
* Cosecant of Angle plus Straight Angle
== Sources ==
* : $\S 5$: Trigonometric Functions: Functions of Angles in All Quadrants in terms of those in Quadrant I
Category:Secant Function
Secant of Angle plus Three Right Angles

== Theorem ==
:$\map \sec {x + \dfrac {3 \pi} 2} = \csc x$
== Proof ==
== Also see ==
* Sine of Angle plus Three Right Angles
* Cosine of Angle plus Three Right Angles
* Tangent of Angle plus Three Right Angles
* Cotangent of Angle plus Three Right Angles
* Cosecant of Angle plus Three Right Angles
== Sources ==
* : $\S 5$: Trigonometric Functions: Functions of Angles in All Quadrants in terms of those in Quadrant I
Category:Secant Function
Secant of Complement equals Cosecant

== Theorem ==
:$\map \sec {\dfrac \pi 2 - \theta} = \csc \theta$ for $\theta \ne n \pi$
where $\sec$ and $\csc$ are secant and cosecant respectively.
That is, the cosecant of an angle is the secant of its complement.
This relation is defined wherever $\sin \theta \ne 0$.
== Proof ==
The above is valid only where $\sin \theta \ne 0$, as otherwise $\dfrac 1 {\sin \theta}$ is undefined.
From Sine of Multiple of Pi it follows that this happens when $\theta \ne n \pi$.
== Examples ==
== Also see ==
* Sine of Complement equals Cosine
* Cosine of Complement equals Sine
* Tangent of Complement equals Cotangent
* Cotangent of Complement equals Tangent
* Cosecant of Complement equals Secant
== Sources ==
* : $\S 5$: Trigonometric Functions: Functions of Angles in All Quadrants in terms of those in Quadrant I
Category:Secant of Complement equals Cosecant
Category:Secant Function
Category:Cosecant Function
Category:Complementary Angles
Secant of Complement equals Cosecant/Examples/3 Theta

== Example of Secant of Complement equals Cosecant ==
:$\map \cos {90 \degrees - 3 \theta} = \csc 3 \theta$
== Proof ==
From Secant of Complement equals Cosecant:
:$\map \cos {90 \degrees - \theta} = \csc \theta$
The result follows by substituting $3 \theta$ for $\theta$.
== Sources ==
* : $\text V$. Trigonometry: Exercise $\text {XXXI}$: $7.$
Category:Secant of Complement equals Cosecant
Secant of Complex Number

== Theorem ==
Let $a$ and $b$ be real numbers.
Let $i$ be the imaginary unit.
Then:
:$\sec \paren {a + b i} = \dfrac {\cos a \cosh b + i \sin a \sinh b} {\cos^2 a \cosh^2 b + \sin^2 a \sinh^2 b}$
where:
:$\sec$ denotes the complex secant function.
:$\sin$ denotes the real sine function
:$\cos$ denotes the real cosine function
:$\sinh$ denotes the hyperbolic sine function
:$\cosh$ denotes the hyperbolic cosine function
== Proof ==
== Also see ==
* Sine of Complex Number
* Cosine of Complex Number
* Tangent of Complex Number
* Cosecant of Complex Number
* Cotangent of Complex Number
Category:Secant Function
Category:Complex Numbers
Secant of Conjugate Angle

== Theorem ==
:$\map \sec {2 \pi - \theta} = \sec \theta$
where $\sec$ denotes secant.
That is, the secant of an angle equals its conjugate.
== Proof ==
== Examples ==
== Also see ==
* Sine of Conjugate Angle
* Cosine of Conjugate Angle
* Tangent of Conjugate Angle
* Cotangent of Conjugate Angle
* Cosecant of Conjugate Angle
== Sources ==
* : $\S 5$: Trigonometric Functions: Functions of Angles in All Quadrants in terms of those in Quadrant I
Category:Secant Function
Category:Conjugate Angles
Secant of Conjugate Angle/Examples/Conjugate of 3 x

== Example of Secant of Conjugate Angle ==
:$\map \sec {360 \degrees - 3 x} = \sec 3 x$
== Proof ==
From Secant of Conjugate Angle:
:$\map \sec {2 \pi - \theta} = \sec \theta$
The result follows by setting $\theta = 3 x$.
== Sources ==
* : $\text V$. Trigonometry: Angles larger than $90 \degrees$: Examples
Category:Secant of Conjugate Angle
Secant of Right Angle

== Theorem ==
:$\sec 90 \degrees = \sec \dfrac \pi 2$ is undefined
where $\sec$ denotes secant.
== Proof ==
From Secant is Reciprocal of Cosine:
:$\sec \theta = \dfrac 1 {\cos \theta}$
From Cosine of Right Angle:
:$\cos \dfrac \pi 2 = 0$
Thus $\sec \theta$ is undefined at this value.
== Also see ==
* Sine of Right Angle
* Cosine of Right Angle
* Tangent of Right Angle
* Cotangent of Right Angle
* Cosecant of Right Angle
== Sources ==
* : $\S 5$: Trigonometric Functions: Exact Values for Trigonometric Functions of Various Angles
Category:Secant Function
Secant of Straight Angle

== Theorem ==
:$\sec 180 \degrees = \sec \pi = -1$
where $\sec$ denotes secant.
== Proof ==
== Also see ==
* Sine of Straight Angle
* Cosine of Straight Angle
* Tangent of Straight Angle
* Cotangent of Straight Angle
* Cosecant of Straight Angle
== Sources ==
* : $\S 5$: Trigonometric Functions: Exact Values for Trigonometric Functions of Various Angles
Category:Secant Function
Secant of Supplementary Angle

== Theorem ==
:$\map \sec {\pi - \theta} = -\sec \theta$
where $\sec$ denotes secant.
That is, the secant of an angle is the negative of its supplement.
== Proof ==
== Also see ==
* Sine of Supplementary Angle
* Cosine of Supplementary Angle
* Tangent of Supplementary Angle
* Cotangent of Supplementary Angle
* Cosecant of Supplementary Angle
== Sources ==
* : $\S 5$: Trigonometric Functions: Functions of Angles in All Quadrants in terms of those in Quadrant I
Category:Secant Function
Category:Supplementary Angles
Secant of Three Right Angles

== Theorem ==
:$\sec 270 \degrees = \sec \dfrac {3 \pi} 2$ is undefined
where $\sec$ denotes secant.
== Proof ==
But from Secant of Right Angle, $\sec 90 \degrees$ is undefined.
== Also see ==
* Sine of Three Right Angles
* Cosine of Three Right Angles
* Tangent of Three Right Angles
* Cotangent of Three Right Angles
* Cosecant of Three Right Angles
== Sources ==
* : $\S 5$: Trigonometric Functions: Exact Values for Trigonometric Functions of Various Angles
Category:Secant Function
Secant of Three Right Angles less Angle

== Theorem ==
:$\map \sec {\dfrac {3 \pi} 2 - \theta} = -\csc \theta$
where $\sec$ and $\csc$ are secant and cosecant respectively.
== Proof ==
== Also see ==
* Sine of Three Right Angles less Angle
* Cosine of Three Right Angles less Angle
* Tangent of Three Right Angles less Angle
* Cotangent of Three Right Angles less Angle
* Cosecant of Three Right Angles less Angle
== Sources ==
* : $\S 5$: Trigonometric Functions: Functions of Angles in All Quadrants in terms of those in Quadrant I
Category:Secant Function
Secant of Zero

== Theorem ==
:$\sec 0 = 0$
where $\sec$ denotes secant.
== Proof ==
== Also see ==
* Sine of Zero is Zero
* Cosine of Zero is One
* Tangent of Zero
* Cotangent of Zero
* Cosecant of Zero
== Sources ==
* : $\S 5$: Trigonometric Functions: Exact Values for Trigonometric Functions of Various Angles
Category:Secant Function
Secant of i

== Theorem ==
:$\sec i = \dfrac {2 e} {e^2 + 1}$
where $\sec$ denotes the complex secant function and $i$ is the imaginary unit.
== Proof 1 ==
== Proof 2 ==
== Also see ==
* Sine of $i$
* Cosine of $i$
* Tangent of $i$
* Cotangent of $i$
* Cosecant of $i$
Category:Secant of i
Category:Secant Function
Category:Imaginary Unit
Second-Countability is Hereditary

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space which is second-countable.
Let $T_H = \struct {H, \tau_H}$, where $\O \subset H \subseteq S$, be a subspace of $T$.
Then $T_H$ is second-countable.
== Proof ==
From the definition of second-countable, $\struct {S, \tau}$ has a countable basis.
That is, $\exists \BB \subseteq \tau$ such that:
:for all $U \in \tau$, $U$ is a union of sets from $\BB$
:$\BB$ is countable.
As $H \subseteq S$ it follows that a $H$ itself is a union of sets from $\BB$.
The result follows from Basis for Topological Subspace‎.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $1$: General Introduction: Countability Properties
* : $\S 3$: New Spaces From Old: Subspaces
Category:Second-Countable Spaces
Category:Topological Subspaces
Second-Countability is Preserved under Open Continuous Surjection

== Theorem ==
Let $T_A = \struct {S_A, \tau_A}$ and $T_B = \struct {S_B, \tau_B}$ be topological spaces.
Let $\phi: T_A \to T_B$ be a surjective open mapping which is also continuous.
If $T_A$ is second-countable, then $T_B$ is also second-countable.
== Proof ==
Let $\phi$ be surjective, continuous and open.
Let $T_A$ be second-countable.
By definition of second-countability $T_A$ has a countable basis, $\BB$, say.
Let $\BB = \set {V_n: n \in \N}$.
We need to show that $\set {\phi \sqbrk {V_n}: n \in \N}$ is a base for $T_B$.
Let $U$ be an open set of $T_B$.
$\phi$ is continuous, so $\phi^{-1} \sqbrk U$ is open in $T_A$.
As $\BB$ is a base for $T_A$, there exists an open set $V_n \subseteq \phi^{-1} \sqbrk U$.
$\phi$ is surjective, so from Surjection iff Right Inverse we have that:
:$\phi \sqbrk {\phi^{-1} \sqbrk U} = U$
So, applying $\phi$ to $V_n$, from Image of Subset under Mapping is Subset of Image we obtain:
:$\phi \sqbrk {V_n} \subseteq U$.
This means that $\set {\phi \sqbrk {V_n}: n \in \N}$ is a base for $T_B$.
Thus, $T_B$ is second-countable.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $3$: Compactness: Invariance Properties
Category:Second-Countable Spaces
Category:Open Mappings
Category:Continuous Mappings
Category:Surjections