nyuuzyou/wikis huggingface dataset:
Tail of Convergent Series tends to Zero

== Theorem ==
Let $\sequence {a_n}_{n \mathop \ge 1}$ be a sequence of real numbers.
Let $\ds \sum_{n \mathop = 1}^\infty a_n$ be a convergent series.
Let $N \in \N_{\ge 1}$ be a natural number.
Let $\ds \sum_{n \mathop = N}^\infty a_n$ be the tail of the series $\ds \sum_{n \mathop = 1}^\infty a_n$.
Then:
:$\ds \sum_{n \mathop = N}^\infty a_n$ is convergent
:$\ds \sum_{n \mathop = N}^\infty a_n \to 0$ as $N \to \infty$.
That is, the tail of a convergent series tends to zero.
== Proof ==
Let $\sequence {s_n}$ be the sequence of partial sums of $\ds \sum_{n \mathop = 1}^\infty a_n$.
Let $\sequence {s'_n}$ be the sequence of partial sums of $\ds \sum_{n \mathop = N}^\infty a_n$.
It will be shown that $\sequence {s'_n}$ fulfils the Cauchy criterion.
That is:
:$\forall \epsilon \in \R_{>0}: \exists N: \forall m, n > N: \size {s'_n - s'_m} < \epsilon$
Let $\epsilon \in \R_{>0}$ be a strictly positive real number.
As $\sequence {s_n}$ is convergent, it conforms to the Cauchy criterion by Convergent Sequence is Cauchy Sequence.
Thus:
:$\exists N: \forall m, n > N: \size {s_n - s_m} < \epsilon$
Now:
and similarly:
:$s_m = s_{N - 1} + s'_m$
Thus:
:$s'_n = s_n - s_{N - 1}$
and:
:$s'_m = s_m - s_{N - 1}$
So:
So $\ds \sum_{n \mathop = N}^\infty a_n$ fulfils the Cauchy criterion.
By Convergent Sequence is Cauchy Sequence it follows that it is convergent.
Now it is shown that $\ds \sum_{n \mathop = N}^\infty a_n \to 0$ as $N \to \infty$.
We have that $\sequence {s_n}$ is convergent,
Let its limit be $l$.
Thus we have:
:$\ds l = \sum_{n \mathop = 1}^\infty a_n = s_{N - 1} + \sum_{n \mathop = N}^\infty a_n$
So:
:$\ds \sum_{n \mathop = N}^\infty a_n = l - s_{N - 1}$
But $s_{N - 1} \to l$ as $N - 1 \to \infty$.
The result follows.
== Also see ==
* Convergent Sequence with Finite Number of Terms Deleted is Convergent
== Sources ==
* : $\S 6.11$
* 
* 
Category:Series
Tail of Decreasing Sequence of Sets is Decreasing

== Theorem ==
Let $X$ be a set.
Let $\sequence {E_n}_{n \mathop \in \N}$ be a decreasing sequence of subsets of $X$.
Then for each $m \in \N$ we have:
:$\sequence {E_{n + m} }_{n \mathop \in \N}$ is a decreasing sequence of sets.
== Proof ==
Since $\sequence {E_n}_{n \mathop \in \N}$ is an decreasing sequence of sets, we have: 
:$E_{n + 1} \subseteq E_n$ for each $n \in \N$.
Swapping $n$ for $n + m$, this in particular gives:
:$E_{n + m + 1} \subseteq E_{n + m}$ for each $n \in \N$.
So $\sequence {E_{n + m} }_{n \mathop \in \N}$ is a decreasing sequence of sets.
Category:Decreasing Sequences of Sets
Tail of Uniformly Convergent Product Converges Uniformly to One

== Theorem ==
Let $X$ be a compact topological space.
Let $\struct {\mathbb K, \norm {\,\cdot\,} }$ be a valued field.
Let $\sequence {f_n}$ be a sequence of continuous mappings $f_n: X \to \mathbb K$.
Let the infinite product $\ds \prod_{n \mathop = 1}^\infty f_n$ converge uniformly on $X$.
Then for all $N \in \N$, $\ds \prod_{n \mathop = N}^\infty f_n$ converges uniformly and the sequence $\ds \prod_{n \mathop = N}^\infty f_n$ converges uniformly to $1$.
== Proof ==
== Also see ==
* Factors in Uniformly Convergent Product Converge Uniformly to One
Category:Uniform Convergence
Category:Infinite Products
Tamref's Last Theorem

== Theorem ==
The Diophantine equation:
:$n^x + n^y = n^z$
has exactly one form of solutions in integers:
:$2^x + 2^x = 2^{x + 1}$
for all $x \in \Z$.
== Proof ==
Since $n^z = n^x + n^y > n^x$ and $n^y$, $z > x,y$.
 assume that $x \le y < z$.
Since both $n^{y - x}$ and $n^{z - y} - 1$ are positive integers, both are equal to $1$.
This gives:
:$y = x$ and $n^{z - y} = 2$
which gives the integer solution:
:$n = 2$, $z - y = 1$
Thus the solutions are:
:$\tuple {n, x, y, z} = \tuple {2, x, x, x + 1}, x \in \Z$
== Historical Note ==
== Sources ==
* : $2$
* : $2$
* : No. $172$
* et al. "tamreF's Last Theorem." Math Fun Facts. <http://www.math.hmc.edu/funfacts>.
Category:Number Theory
Tamref's Last Theorem/Mistake

== Source Work ==
:
:The Dictionary
::$2$
:
:The Dictionary
::$2$
== Mistake ==
:Fermat's equation being exceedingly difficult to solve, several mathematicians have noticed in an idle moment that $n^x + n^y = n^z$ is much easier. Its only solutions in integers are when $n = 2$, and $2^1 + 2^1 = 2^2$.
== Correction ==
In fact $2^x + 2^y = 2^z$ has solutions for all $x \in \Z$, whence $y = x$ and $z = x + 1$, for example:
== Sources ==
* : $2$
* : $2$
Category:Mistakes/Curious and Interesting Numbers
Tamura-Kanada Circuit Method

== Algorithm ==
The following algorithm can be used to calculate $\pi$ (pi):
:$(1): \quad$ Set $A = X = 1$, set $B = \dfrac 1 {\sqrt 2}$, and set $C = \dfrac 1 4$
:$(2): \quad$ Set $Y = A$.
:$(3): \quad$ Set $A = \dfrac {A + B} 2$
:$(4): \quad$ Set $B = \sqrt {B Y}$.
:$(5): \quad$ Set $C = C - X \paren {A - Y}^2$
:$(6): \quad$ Set $X = 2 X$
:$(7): \quad$ Output $\dfrac {\paren {A + B}^2} {4 C}$ as an approximation to $\pi$.
:$(8): \quad$ For a better approximation to $\pi$, set $Y$ equal to the output, return to step $(2)$ and continue.
=== Example ===
== Proof ==
== Historical Note ==
== Sources ==
* : $3 \cdotp 14159 \, 26535 \, 89793 \, 23846 \, 26433 \, 83279 \, 50288 \, 41972 \ldots$
* : $3 \cdotp 14159 \, 26535 \, 89793 \, 23846 \, 26433 \, 83279 \, 50288 \, 41971 \ldots$
Category:Approximations to Pi
Tamura-Kanada Circuit Method/Example

== Example of Use of Tamura-Kanada Circuit Method ==
Starting with $A = X = 1$, $B = \dfrac 1 {\sqrt 2}$, $C = \dfrac 1 4$, the successive values of $\dfrac {\paren {A + B}^2} {4 C}$ on the first $3$ loops are:
and it is seen that the value for $\pi$ is already correct to $6$ decimal places.
== Sources ==
* : $3 \cdotp 14159 \, 26535 \, 89793 \, 23846 \, 26433 \, 83279 \, 50288 \, 41972 \ldots$
* : $3 \cdotp 14159 \, 26535 \, 89793 \, 23846 \, 26433 \, 83279 \, 50288 \, 41971 \ldots$
Category:Approximations to Pi
Tamura-Kanada Circuit Method/Example/Mistake

== Source Work ==
:
:The Dictionary
::$3 \cdotp 14159 \, 26535 \, 89793 \, 23846 \, 26433 \, 83279 \, 50288 \, 41972 \ldots$
== Mistake ==
:Here are the values of $\pi$ after going round just $3$ times on a pocket calculator. It is already correct to $5$ decimal places!
It is in fact seen that the value for $\pi$ is actually correct to $6$ decimal places, not $5$.
In , this has been corrected.
== Sources ==
* : $3 \cdotp 14159 \, 26535 \, 89793 \, 23846 \, 26433 \, 83279 \, 50288 \, 41972 \ldots$
Category:Mistakes/Curious and Interesting Numbers
Tamura-Kanada Circuit Method/Historical Note

== Historical Note on Tamura-Kanada Circuit Method ==
This algorithm was used by , and to calculate $\pi$ (pi) to $16 \, 777 \, 216$ digits in $1983$.
== Sources ==
* : $3 \cdotp 14159 \, 26535 \, 89793 \, 23846 \, 26433 \, 83279 \, 50288 \, 41972 \ldots$
Category:Approximations to Pi
Tangent-Chord Theorem

== Theorem ==
Let $EF$ be a tangent to a circle $ABCD$, touching it at $B$.
Let $BD$ be a chord of $ABCD$.
Then:
:the angle in segment $BCD$ equals $\angle DBE$
and: 
:the angle in segment $BAD$ equals $\angle DBF$.
== Proof ==
:
Draw $BA$ perpendicular to $EF$ through $B$.
Let $C$ be selected on the circle on the arc $BD$.
Join $AD, DC, CB$.
From Right Angle to Tangent of Circle goes through Center, the center of the circle lies on $AB$.
By definition, then, $AB$ is a diameter of the circle.
From Relative Sizes of Angles in Segments, it follows that $\angle ADB$ is a right angle.
Therefore from Sum of Angles of Triangle Equals Two Right Angles $\angle BAD + \angle ABD$ equals a right angle.
But $\angle ABF$ is also a right angle.
So $\angle ABF = \angle BAD + \angle ABD$
Subtracting $\angle ABD$ from each, it follows that $\angle DBF = \angle BAD$.
Next we have that $ABCD$ is a cyclic quadrilateral.
From Opposite Angles of Cyclic Quadrilateral sum to Two Right Angles $\angle BAD + \angle BCD$ equals two right angles.
But from Two Angles on Straight Line make Two Right Angles, so does $\angle DBE + \angle DBF$.
But as $\angle BAD = \angle DBF$ it follows that $\angle BCD = \angle DBE$.
== Also known as ==
== Sources ==
* : Book $\text{III}$. Propositions
* $(7)$
* 
Category:Tangent-Chord Theorem
Category:Circles
Category:Tangents to Circles
Category:Named Theorems
Tangent Function is Odd

== Theorem ==
For all $x \in \C$ where $\tan x$ is defined:
:$\map \tan {-x} = -\tan x$
That is, the tangent function is odd.
== Proof ==
== Also see ==
* Sine Function is Odd
* Cosine Function is Even
* Cotangent Function is Odd
* Secant Function is Even
* Cosecant Function is Odd
* Hyperbolic Tangent Function is Odd
== Sources ==
* : $\S 5$: Trigonometric Functions: $5.30$
* : $\S 5$: Trigonometric Functions: Functions of Angles in All Quadrants in terms of those in Quadrant I
* : $2$: Functions, Limits and Continuity: The Elementary Functions: $4$
* 
* 
* : Appendix $12$: Trigonometric formulae: Symmetry
* : Appendix $14$: Trigonometric formulae: Symmetry
Category:Tangent Function
Category:Examples of Odd Functions
Tangent Function is Periodic on Reals

== Theorem ==
The real tangent function is periodic with period $\pi$.
This can be written:
:$\tan x = \map \tan {x \bmod \pi}$
where $x \bmod \pi$ denotes the modulo operation.
== Proof ==
From Derivative of Tangent Function, we have that:
:$\map {D_x} {\tan x} = \dfrac 1 {\cos^2 x}$
provided $\cos x \ne 0$.
From Shape of Cosine Function, we have that $\cos > 0$ on the interval $\openint {-\dfrac \pi 2} {\dfrac \pi 2}$.
From Derivative of Monotone Function, $\tan x$ is strictly increasing on that interval, and hence can not have a period of less than $\pi$.
Hence the result.
== Sources ==
* : $\S 16.5 \ (2)$
* : $\S 1.2.4$: Integer Functions and Elementary Number Theory
* : Appendix $12$: Trigonometric formulae: Shifts and periodicity
* : Appendix $14$: Trigonometric formulae: Shifts and periodicity
Category:Tangent Function
Tangent Half-Angle Substitution for Cosine

== Corollary to Double Angle Formula for Cosine ==
:$\cos 2 \theta = \dfrac {1 - \tan^2 \theta} {1 + \tan^2 \theta}$
where $\cos$ and $\tan$ denote cosine and tangent respectively.
== Proof ==
== Sources ==
* : $\text V$. Trigonometry: Formulae $(27)$
Category:Tangent Half-Angle Substitution for Cosine
Category:Cosine Function
Category:Tangent Function
Category:Double Angle Formula for Cosine
Tangent Half-Angle Substitution for Sine

== Corollary to Double Angle Formula for Sine ==
:$\sin 2 \theta = \dfrac {2 \tan \theta} {1 + \tan^2 \theta}$
where $\sin$ and $\tan$ denote sine and tangent respectively.
== Proof ==
== Sources ==
* : $\text V$. Trigonometry: Formulae $(26)$
Category:Tangent Half-Angle Substitution for Sine
Category:Sine Function
Category:Tangent Function
Category:Double Angle Formula for Sine
Tangent Inequality

== Theorem ==
:$x < \tan x$
for all $x$ in the interval $\openint 0 {\dfrac \pi 2}$.
== Proof ==
Let $\map f x = \tan x - x$.
By Derivative of Tangent Function, $\map {f'} x = \sec^2 x - 1$.
By Shape of Secant Function, $\sec^2 x > 1$ for $x \in \openint 0 {\dfrac \pi 2}$.
Hence $\map {f'} x > 0$.
From Derivative of Monotone Function, $\map f x$ is strictly increasing in this interval.
Since $\map f 0 = 0$, it follows that $\map f x > 0$ for all $x \in \openint 0 {\dfrac \pi 2}$.
Category:Tangent Function
Category:Inequalities
Tangent Line to Convex Graph

== Theorem ==
Let $f$ be a real function that is:
:continuous on some closed interval $\closedint a b$
:differentiable and convex on the open interval $\openint a b$. 
Then all the tangent lines to $f$ are below the graph of $f$.
== Proof ==
:
Let $\TT$ be the tangent line to $f$ at some point $\tuple {c, \map f c}$, $c \in \openint a b$.
Let the gradient of $\TT$ be $m$.
Let $\tuple {x_1, y_1}$ be an arbitrary point on $\TT$.
From the point-slope form of a straight line:
For $\TT$:
:$y = \map \TT x$
:$y_1 = \map f c$
:$m = \map {f'} c$
:$x = x$
:$x_1 = c$
so:
:$\map \TT x = \map {f'} c \paren {x - c} + \map f c$
Consider the graph of $f$ to the right of $\tuple {c, \map f c}$, that is, any $x$ in $\openint c b$. 
Let $d$ be the directed vertical distance from $\TT$ to the graph of $f$.
That is, if $f$ is above $\TT$ then $d > 0$.
If $f$ is below $\TT$, then $d < 0$. 
(From the diagram, it is apparent that $\TT$ is below $f$, but we shall prove it analytically.)
$d$ can be evaluated by:
By the Mean Value Theorem, there exists some constant $k$ in $\openint c b$ such that:
Substitute this into the formula for $d$:
Recall that $x$ lies in the interval $\openint c b$.
So $x > c$, and the quantity $x - c$ is (strictly) positive.
$k$ is also in the interval $\openint c b$ and so $k > c$.
By construction, $f$ is convex. 
By the definition of convex:
:$k > c \implies \map {f'} k > \map {f'} c$
which means that:
:$\paren {\map {f'} k - \map {f'} c} > 0$
Then $d$ is the product of two (strictly) positive quantities and is itself (strictly) positive. 
Similarly, consider the graph of $f$ to the left of $\tuple {c, \map f c}$, that is, any $x$ in $\openint a c$.
By the same process as above, we will have:
:$d = \paren {\map {f'} k - \map {f'} c} \paren {x - c}$
This time, $x < c$ and the quantity $x - c$ is (strictly) negative.
Further, $k < c$, and so by a similar argument as above:
:$k < c \implies \map {f'} k < \map {f'} c$
and the quantity $\paren {\map {f'} k - \map {f'} c}$ is also (strictly) negative.
Thus $d$ will be the product of two (strictly) negative quantities, and will again be (strictly) positive. 
== Also see ==
* Tangent Line to Concave Graph
* Tangent Line at Inflection Point
== Sources ==
* : Appendix $A$: Concavity Interpretation
Category:Differential Calculus
Category:Analytic Geometry
Category:Convex Real Functions
Category:Tangents
Tangent Plane to Conical Surface touches along Straight Line

== Theorem ==
Let $\CC$ be a conical surface.
Let $\TT$ be a tangent plane to $\CC$.
Then the element of contact between $\CC$ and $\TT$ is a straight line.
== Proof ==
== Sources ==
* 
* 
Category:Elements of Contact
Category:Tangent Planes
Category:Cones
Tangent Plane to Cylindrical Surface touches along Straight Line

== Theorem ==
Let $\CC$ be a cylindrical surface.
Let $\TT$ be a tangent plane to $\CC$.
Then the element of contact between $\CC$ and $\TT$ is a straight line.
== Proof ==
== Sources ==
* 
* 
Category:Elements of Contact
Category:Tangent Planes
Category:Cylinders
Tangent Points of Incircle in Terms of Semiperimeter

== Theorem == 
Let $\triangle ABC$ be a triangle whose sides are $a$, $b$ and $c$ opposite vertices $A$, $B$ and $C$ respectively.
Let $s$ denote the semiperimeter of $\triangle ABC$.
Then the distance from $A$ to a point tangent to the incircle is equal to $s - a$.
== Proof ==
Let $D$, $E$ and $F$ be the points where the incircle is tangent to the sides $AC$, $AB$ and $BC$ respectively.
Then:
The above argument can be used to show that:
Category:Incircles of Triangles
Tangent Secant Theorem

== Theorem ==
Let $D$ be a point outside a circle $ABC$.
Let $DB$ be tangent to the circle $ABC$.
Let $DA$ be a straight line which cuts the circle $ABC$ at $A$ and $C$.
Then $DB^2 = AD \cdot DC$.
== Proof 1 ==
== Proof 2 ==
== Also known as ==
== Also see ==
* Secant Secant Theorem
== Sources ==
* $(5)$
* $(5)$
Category:Tangent Secant Theorem
Category:Circles
Category:Tangents to Circles
Category:Named Theorems
Tangent Secant Theorem/Proof 1

== Theorem ==
== Proof ==
Let $DA$ pass through the center $F$ of circle $ABC$.
Join $FB$.
From Radius at Right Angle to Tangent, $\angle FBD$ is a right angle.
:
We have that $F$ bisects $AC$ and that $CD$ is added to it.
So we can apply Square of Sum less Square and see that:
:$AD \cdot DC + FC^2 = FD^2$
But $FC = FB$ and so:
:$AD \cdot DC + FB^2 = FD^2$
But from Pythagoras's Theorem we have that $FD^2 = FB^2 + DB^2$ and so:
:$AD \cdot DC + FB^2 = FB^2 + DB^2$
from which it follows that:
:$AD \cdot DC = DB^2$
which is what we wanted to show.
Now let $DA$ be such that it does not pass through the center $E$ of circle $ABC$.
Draw $EF$ perpendicular to $DA$ and draw $EB, EC, ED$.
:
From Radius at Right Angle to Tangent, $\angle EBD$ is a right angle.
From Conditions for Diameter to be Perpendicular Bisector, $EF$ bisects $AC$.
So $AF = FC$.
So we can apply Square of Sum less Square and see that:
:$AD \cdot DC + FC^2 = FD^2$
Let $FE^2$ be added to each:
:$AD \cdot DC + FC^2 + FE^2 = FD^2 + FE^2$
Now $\angle DFE$ is a right angle and so by Pythagoras's Theorem we have:
:$FD^2 + FE^2 = ED^2$
:$FC^2 + FE^2 = EC^2$
This gives us:
:$AD \cdot DC + EC^2 = ED^2$
But $EC = EB$ as both are radii of the circle $ABC$.
Next note that $\angle EBD$ is a right angle and so by Pythagoras's Theorem we have:
:$ED^2 = EB^2 + DB^2$
which gives us:
:$AD \cdot DC + EB^2 = EB^2 + DB^2$
from which it follows that:
:$AD \cdot DC = DB^2$
which is what we wanted to show.
== Sources ==
* : Book $\text{III}$. Propositions
Category:Tangent Secant Theorem
Tangent Secant Theorem/Proof 2

== Theorem ==
== Proof ==
Let the circle $\CC$ be embedded in a Cartesian plane with its center located at the origin.
Let $\CC$ have radius $r$.
From Equation of Circle center Origin, $\CC$ can be described as:
:$(1): \quad x^2 + y^2 = r^2$
From Equation of Straight Line in Plane: Parametric Form, let the straight line $DCA$ be described using the parametric equations:
:$\begin {cases} x = x_0 + \rho \cos \psi \\ y = y_0 + \rho \sin \psi \end {cases}$
where $D = \tuple {x_0, y_0}$.
Substituting these for $x$ and $y$ into equation $(1)$:
We see that $(2)$ is a quadratic equation in $\rho$.
Let $\rho_1$ and $\rho_2$ be the solutions of $(2)$.
Each of these corresponds to one of the points at which $DCA$ intersects $\CC$, where:
:$\rho_1 = DA$
:$\rho_2 = DC$
From Product of Roots of Quadratic Equation:
:$\rho_1 \rho_2 = x_0^2 + y_0^2 - r^2$
From Length of Tangent from Point to Circle center Origin:
:$DB = x_0^2 + y_0^2 - r^2$
Hence we have:
:$DA \cdot DC = DB^2$
== Sources ==
* : Chapter $\text {III}$. The Circle: $11$. Geometrical meaning of the expression $x^2 + y^2 - r^2$
Category:Tangent Secant Theorem
Tangent Space as Orthogonal Direct Sum of Horizontal and Vertical Tangent Spaces

== Theorem ==
Let $M$ be a Riemannian manifold.
Let $x \in M$ be a point.
Let $H_x$ and $V_x$ be a horizontal and vertical tangent space of $M$ at $x$ respectively.
Let $T_x M$ be the tangent space of $M$ at $x$.
Then $T_x M$ decomposes as the orthogonal direct sum of $H_x$ and $V_x$:
:$T_x M = H_x \oplus V_x$
== Proof ==
== Sources ==
* : $\S 2$: Riemannian Metrics. Methods for Constructing Riemannian Metrics
Category:Riemannian Geometry
Tangent Space is Vector Space

== Theorem ==
Let $M$ be a smooth manifold of dimension $n \in \N$. 
Let $m \in M$ be a point. 
Let $\struct {U, \kappa}$ be a chart with $m \in U$. 
Let $T_m M$ be the tangent space at $m$. 
Then $T_m M$ is a real vector space of dimension $n$, spanned by the basis: 
:$\set {\valueat {\dfrac \partial {\partial \kappa^i} } m : i \in \set {1, \dotsc, n} }$
that is, the set of partial derivatives with respect to the $i$th coordinate function $\kappa^i$ evaluated at $m$.
== Proof ==
Let $V$ be an open neighborhood of $m$ with $V \subseteq U \subseteq M$. 
Let $\map {C^\infty} {V, \R}$ be the set of smooth mappings $f: V \to \R$. 
Let $X_m, Y_m \in T_m M$.
Let $\lambda \in \R$.
Then, by definition of tangent vector and Equivalence of Definitions of Tangent Vector:
:$X_m, Y_m$ are linear transformations on $\map {C^\infty} {V, \R}$. 
Hence $\paren {X_m + \lambda Y_m}$ are also linear transformations. 
Therefore, it is enough to show that $X_m + \lambda Y_m$ satisfies the Leibniz law. 
Let $f, g \in \map {C^\infty} {V, \R}$. 
Then:
It follows that:
:$X_m + \lambda Y_m \in T_m M$
Hence $T_m M$ is a real vector space. 
Again, by definition of tangent vector and Equivalence of Definitions of Tangent Vector:
:for all $X_m \in T_m M$ there exists a smooth curve:
::$\gamma: I \subseteq \R \to M$
:where $\map \gamma 0 = m$ such that:
We define:
:$X^i_m := \map {\dfrac {\map \d {\kappa^i \circ \gamma} } {\d \tau} } 0$
and as above: 
:$\valueat {\map {\dfrac \partial {\partial \kappa^i} } m} f := \map {\dfrac {\map \partial {f \circ \kappa^{-1} } } {\partial \kappa^i} } {\map \kappa m}$
Therefore: 
:$\ds \map {X_m} f = \map {\paren {\sum_{i \mathop = 1}^n X^i_m \valueat {\dfrac \partial {\partial \kappa^i} } m} } f$
:
:$\ds X_m = \sum_{i \mathop = 1}^n X^i_m \valueat {\frac \partial {\partial \kappa^i} } m$
Hence:
:$\set {\valueat {\dfrac \partial {\partial \kappa^i} } m: i \in \set {1, \dotsc, n} }$
forms a basis. 
Hence, by definition of dimension of vector space:
:$\dim T_m M = n = \dim M$
This completes the proof.
Tangent and Cotangent are Cofunctions

== Theorem ==
The tangent and cotangent are cofunctions:
== Proof ==
=== Proof of $(1)$ ===
From Tangent is Sine divided by Cosine:
:$\tan x = \dfrac {\sin x} {\cos x}$
Hence in order for $\tan x$ to be defined it is necessary for $\cos x \ne 0$.
Then we have:
:Tangent of Complement equals Cotangent
=== Proof of $(2)$ ===
From Cotangent is Cosine divided by Sine:
:$\cot x = \dfrac {\cos x} {\sin x}$
Hence in order for $\cot x$ to be defined it is necessary for $\sin x \ne 0$.
Then we have:
:Cotangent of Complement equals Tangent
Hence the result by definition of cofunction.
== Sources ==
* 
* 
Category:Cofunctions
Tangent in terms of Hyperbolic Tangent

== Theorem ==
Let $z \in \C$ be a complex number.
Then:
:$i \tan z = \map \tanh {i z}$
where:
:$\tan$ denotes the tangent function
:$\tanh$ denotes the hyperbolic tangent
:$i$ is the imaginary unit: $i^2 = -1$.
== Proof ==
 
== Also see ==
* Sine in terms of Hyperbolic Sine
* Cosine in terms of Hyperbolic Cosine
* Cotangent in terms of Hyperbolic Cotangent
* Secant in terms of Hyperbolic Secant
* Cosecant in terms of Hyperbolic Cosecant
== Sources ==
* : $\S 8$: Hyperbolic Functions: $8.82$: Relationship between Hyperbolic and Trigonometric Functions
* : $2$: Functions, Limits and Continuity: The Elementary Functions: $5$
* 
* 
Category:Tangent Function
Category:Hyperbolic Tangent Function
Tangent in terms of Secant

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

== Theorem ==
Let $\theta$ be an angle such that $\sin \theta \ne 0$ and $\cos \theta \ne 0$.
Then:
:$\tan \theta = \dfrac 1 {\cot \theta}$
where $\tan$ denotes the tangent function and $\cot$ denotes the cotangent function.
== Proof ==
$\tan \theta$ is not defined when $\cos \theta = 0$, and $\cot \theta$ is not defined when $\sin \theta = 0$.
== Also see ==
* Trigonometric Functions in terms of each other
Category:Tangent Function
Category:Cotangent Function
Tangent is Sine divided by Cosine

== Theorem ==
Let $\theta$ be an angle such that $\cos \theta \ne 0$.
Then:
:$\tan \theta = \dfrac {\sin \theta} {\cos \theta}$
where $\tan$, $\sin$ and $\cos$ mean tangent, sine 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 $\cos \theta = 0$ the expression $\dfrac {\sin \theta} {\cos \theta}$ is not defined.
== Sources ==
* : $\text V$. Trigonometry: Definitions of the ratios
* : $\S 5$: Trigonometric Functions: $5.15$
* 
* 
* : Appendix $12$: Trigonometric formulae
* : Appendix $14$: Trigonometric formulae
Category:Sine Function
Category:Cosine Function
Category:Tangent Function
Tangent of 105 Degrees

== Theorem ==
:$\tan 105^\circ = \tan \dfrac {7 \pi} {12} = - \left({2 + \sqrt 3}\right)$
where $\tan$ denotes tangent.
== Proof ==
== Sources ==
* : $\S 5$: Trigonometric Functions: Exact Values for Trigonometric Functions of Various Angles
Category:Tangent Function
Tangent of 120 Degrees

== Theorem ==
:$\tan 120 \degrees = \tan \dfrac {2 \pi} 3 = -\sqrt 3$
where $\tan$ denotes tangent.
== Proof ==
== Sources ==
* : $\S 5$: Trigonometric Functions: Exact Values for Trigonometric Functions of Various Angles
Category:Tangent Function