hcju/proofwikiqa huggingface dataset:
Proof by induction: === Basis for the Induction === $n = 2$ holds by Determinant of Matrix Product. So shown for base case. === Induction Hypothesis === This is our induction hypothesis: :$\map \det {\mathbf A_1 \mathbf A_2 \cdots \mathbf A_k} = \map \det {\mathbf A_1} \map \det {\mathbf A_2} \cdots \map \det {\mathbf A_k}$ Now we need to show true for $n = k + 1$: :$\map \det {\mathbf A_1 \mathbf A_2 \cdots \mathbf A_k \mathbf A_{k + 1} } = \map \det {\mathbf A_1} \map \det {\mathbf A_2} \cdots \map \det {\mathbf A_k} \map \det {\mathbf A_{k + 1} }$ === Induction Step === This is our induction step: {{begin-eqn}} {{eqn | l = \map \det {\mathbf A_1 \mathbf A_2 \cdots \mathbf A_k \mathbf A_{k + 1} }       | r = \map \det {\mathbf A_1 \mathbf A_2 \cdots \mathbf A_k} \map \det {\mathbf A_{k + 1} }       | c = Determinant of Matrix Product }} {{eqn | r = \map \det {\mathbf A_1} \map \det {\mathbf A_2} \cdots \map \det {\mathbf A_k} \map \det {\mathbf A_{k + 1} }       | c = Induction hypothesis }} {{end-eqn}} The result follows by induction: :$\map \det {\mathbf A_1 \mathbf A_2 \cdots \mathbf A_n} = \map \det {\mathbf A_1} \map \det {\mathbf A_2} \cdots \map \det {\mathbf A_n}$ {{qed}} Category:Determinant of Matrix Product gcka4srqc7leplzrelff1nhk7zlsiq3
Join $AC$. {{begin-eqn}} {{eqn | n = 1       | l = AB       | r = CD       | c = Given }} {{eqn | n = 2       | l = BC       | r = DA       | c = Given }} {{eqn | n = 3       | l = AC       | r = CA       | c = Equality is Reflexive }} {{eqn | n = 4       | l = \Delta ABC       | r = \Delta CDA       | c = SSS from $(1)$, $(2)$, and $(3)$ }} {{eqn | n = 5       | l = \angle BCA       | r = \angle DAC       | c = from $(4)$ }} {{eqn | n = 6       | l = \angle BAC       | r = \angle DCA       | c = from $(4)$ }} {{eqn | n = 7       | l = BC       | o = \parallel       | r = DA       | c = Equal Alternate Angles implies Parallel Lines from $(5)$ }} {{eqn | n = 8       | l = BA       | o = \parallel       | r = DC       | c = Equal Alternate Angles implies Parallel Lines from $(6)$ }} {{end-eqn}} From $(7)$ and $(8)$, it follows by definition that $ABCD$ is a parallelogram. {{qed}} Category:Parallelograms bb5ds8cs25ff241o6a6jcnghhw6kv8i
Let $S_p$ be the set of all elements of $S$ preceding $p$: :$S_p = \set {x \in S: x \prec p}$ Let $T' = T \cup S_p$. Then the set $T'$ satisfies the conditions of the Principle of Mathematical Induction for a Naturally Ordered Semigroup. From that result: :$T' = S$ By Set Difference with Union is Set Difference: :$S \setminus S_p = T \setminus S_p$ By Set Difference is Subset: :$T \setminus S_p \subseteq T$ completing the proof. {{qed}}
The hypotheses state precisely that $S$ is an infinite successor set. Then the minimal infinite successor set $\omega$ being defined as the intersection of all infinite successor sets, we conclude that: :$\omega \subseteq S$ by Intersection is Subset: General Result. Thus, by definition of set equality: :$S = \omega$ {{qed}}
By definition of the natural numbers in $\R$: :$\N = \displaystyle \bigcap \II$ where $\II$ is the set of inductive sets in $\R$. Since $A$ was supposed to be inductive, it follows that: :$\N \subseteq A$ from Intersection is Subset: General Result. Hence by definition of set equality: :$A = \N$ {{qed}}
The mapping $\phi$ is indeed a ring homomorphism, because each $A \to A / I_i$ is a ring homomorphism. The kernel of $\phi$ is given by: :$\displaystyle \ker \phi = \left\{{x \in A: \forall i, 1 \le i \le k : x \in I_i}\right\} = \bigcap_{1 \mathop \le i \mathop \le k} I_i$ So $\phi$ defines an injective homomorphism by passing through the quotient. It remains then to prove that: :$\tilde \phi \equiv \phi \circ \pi$ where $\pi$ is the canonical quotient-map, is surjective. That is: :$\forall x_i \in A, 1 \le i \le k: \exists x \in A: x - x_i \in I_i, 1 \le i \le k$ Then it follows that: :$\displaystyle \left({x_1 + I_1, \ldots, x_k + I_k}\right) = \tilde \phi \left({x + \bigcap I_i}\right)$ Note that: :$\left({x_1 + I_1, \ldots, x_k + I_k}\right) = \left({x_1 + I_1}\right) e_1 + \cdots + \left({x_k + I_k}\right) e_k$ where the unit element lies at the $i$-th component: :$e_i = \left({0, \ldots, 0, 1_{A / I_i}, 0, \ldots, 0}\right)$ {{explain|Unity not unit, yeah?}} This implies that it is enough to find $a_i \in A, 1 \le i \le k$, such that: :$\tilde \phi (a_i + \bigcap I_i) = e_i$ since: {{begin-eqn}} {{eqn | l = \tilde \phi \left({x_1 a_1 + \cdots + x_k a_k}\right)       | r = \tilde \phi \left({x_1}\right) \tilde \phi \left({a_1}\right) + \cdots + \tilde \phi \left({x_k}\right) \tilde \phi \left({a_k}\right)       | c =  }} {{eqn | r = \left({\tilde \phi \left({x_1}\right), \ldots, \tilde \phi \left({x_k}\right)}\right)       | c =  }} {{eqn | r = \left({x_1 + I_1, \ldots, x_n + I_n}\right)       | c =  }} {{end-eqn}} We show that $e_1$ is in the image of $\tilde \phi$. The other instances of $e_i$ then follow in a similar manner. We then need to find an $a \in A$ such that $a -1 \in I_1, a \in I_2, \ldots, a \in I_k$. Since then $I_1$ is coprime with the other ideals, we have that $I_1 + I_j = A, 2 \le j \le k$. So there exists a $b_j \in I_1$, $c_j \in I_j$ such that $b_j + c_j = 1$. Note that this implies that $c_j - 1 \in I_1, c_j \in I_j$.  Define now $a = c_2 c_3 \cdots c_k \in A$. Then for $2 \le j \le k$: :$a = c_j \left({c_2 \cdots c_{j - 1} c_{j + 1} \cdots c_k}\right) \in I_j$ and: {{begin-eqn}} {{eqn | l = a + I_1       | r = \left({c_2 + I_1}\right) \cdots \left({c_k + I_1}\right)       | c =  }} {{eqn | r = \left({1 + I_1}\right) \cdots \left({1 + I_1}\right)       | c =  }} {{eqn | r = 1 + I_1       | c =  }} {{end-eqn}} Hence: :$a - 1 \in I_1$ {{qed}} Category:Commutative Algebra 38mf2x0o58risu8v5x45d5bltmq92o1
We have: {{begin-eqn}} {{eqn | n = 1       | l = \cos i + i \sin i       | r = e^{i \times i}       | c = Euler's Formula }} {{eqn | r = e^{-1}       | c = Definition of Imaginary Unit }} {{eqn | r = \frac 1 e }} {{end-eqn}} Also: {{begin-eqn}} {{eqn | n = 2       | l = \cos i - i \sin i       | r = \cos \left({-i}\right) + i \sin \left({-i}\right)       | c = Cosine Function is Even and Sine Function is Odd }} {{eqn | r = e^{i \times \left({-i}\right)}       | c = Euler's Formula }} {{eqn | r = e^1       | c = Definition of Imaginary Unit }} {{eqn | r = e }} {{end-eqn}} Then from $(1) - (2)$: {{begin-eqn}} {{eqn | l = 2 i \sin i       | r = \frac 1 e - e }} {{eqn | ll= \implies       | l = \sin i       | r = \frac 1 {2 i} \left({\frac 1 e - e}\right) }} {{eqn | r = \left({\frac e 2 - \frac 1 {2 e} }\right) i }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \sin i       | r = i \sinh 1       | c = Hyperbolic Sine in terms of Sine }} {{eqn | r = i \frac {e^1 - e^{-1} } 2       | c = {{Defof|Hyperbolic Sine}} }} {{eqn | r = \paren {\frac e 2 - \frac 1 {2 e} } i }} {{end-eqn}} {{qed}}
We have: {{begin-eqn}} {{eqn | n = 1       | l = \cos i + i \sin i       | r = e^{i \times i}       | c = Euler's Formula }} {{eqn | r = e^{-1}       | c = Definition of Imaginary Unit }} {{eqn | r = \frac 1 e }} {{end-eqn}} Also: {{begin-eqn}} {{eqn | n = 2       | l = \cos i - i \sin i       | r = \cos \left({-i}\right) + i \sin \left({-i}\right)       | c = Cosine Function is Even and Sine Function is Odd }} {{eqn | r = e^{i \times \left({-i}\right)}       | c = Euler's Formula }} {{eqn | r = e^1       | c = Definition of Imaginary Unit }} {{eqn | r = e }} {{end-eqn}} Then from $(1) + (2)$: {{begin-eqn}} {{eqn | l = 2 \cos i       | r = \frac 1 e + e }} {{eqn | ll= \implies       | l = \cos i       | r = \frac 1 2 \left({\frac 1 e + e}\right) }} {{eqn | r = \frac e 2 + \frac 1 {2 e} }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \cos i       | r = \cosh 1       | c = Hyperbolic Cosine in terms of Cosine }} {{eqn | r = \frac {e^1 + e^{-1} } 2       | c = {{Defof|Hyperbolic Cosine}} }} {{eqn | r = \frac e 2 + \frac 1 {2 e} }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \tan i       | r = i \tanh 1       | c = Hyperbolic Tangent in terms of Tangent }} {{eqn | r = \paren {\frac {e^1 - e^{-1} } {e^1 + e^{-1} } } i       | c = {{Defof|Hyperbolic Tangent}} }} {{eqn | r = \paren {\frac {e^2 - 1} {e^2 + 1} } i       | c = multiplying denominator and numerator by $e$ }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \tan i       | r = \frac {\sin i} {\cos i}       | c = {{Defof|Complex Tangent Function}} }} {{eqn | r = \frac {\left({\frac e 2 - \frac 1 {2 e} }\right) i} {\frac e 2 + \frac 1 {2 e} }       | c = Sine of $i$ and Cosine of $i$ }} {{eqn | r = \left({\frac {e - \frac 1 e } {e + \frac 1 e} }\right) i       | c = multiplying denominator and numerator by $2$ }} {{eqn | r = \left({\frac {e^2 - 1} {e^2 + 1} }\right) i       | c = multiplying denominator and numerator by $e$ }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \csc i       | r = -i \csch 1       | c = Hyperbolic Cosecant in terms of Cosecant }} {{eqn | r = -\paren {\frac 2 {e^1 - e^{-1} } } i       | c = {{Defof|Hyperbolic Cosecant}} }} {{eqn | r = -\paren {\frac {2 e} {e^2 - 1} } i       | c = multiplying denominator and numerator by $e$ }} {{eqn | r = \paren {\frac {2 e} {1 - e^2} } i }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \csc i       | r = \frac 1 {\sin i}       | c = Definition of Complex Cosecant }} {{eqn | r = \frac 1 {\left({ \frac e 2 - \frac 1 {2e} }\right) i}       | c = Sine of $i$ }} {{eqn | r = \left({ \frac 1 {\frac 1 {2e} - \frac e {2} } }\right) i       | c = Reciprocal of $i$ }} {{eqn | r = \left({\frac {2 e} {1 - e^2} }\right) i       | c = multiplying denominator and numerator by $2 e$ }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \sec i       | r = \frac 1 {\cos i}       | c = Definition of Complex Secant }} {{eqn | r = \frac 1 {\frac e 2 + \frac 1 {2 e} }       | c = Cosine of $i$ }} {{eqn | r = \frac {2 e} {e^2 + 1}       | c = multiplying denominator and numerator by $2 e$ }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \sec i       | r = \sech 1       | c = Hyperbolic Secant in terms of Secant }} {{eqn | r = \frac 2 {e^1 + e^{-1} }       | c = {{Defof|Hyperbolic Secant}} }} {{eqn | r = \frac {2 e} {e^2 + 1}       | c = multiplying denominator and numerator by $e$ }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \cot i       | r = -i \coth 1       | c = Hyperbolic Cotangent in terms of Cotangent }} {{eqn | r = -\paren {\frac {e^1 + e^{-1} } {e^1 - e^{-1} } } i       | c = {{Defof|Hyperbolic Cotangent|index = 1}} }} {{eqn | r = -\paren {\frac {e^2 + 1} {e^2 - 1} } i       | c = multiplying denominator and numerator by $e$ }} {{eqn | r = \paren {\frac {1 + e^2} {1 - e^2} } i }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \cot i       | r = \frac {\cos i} {\sin i}       | c = {{Defof|Complex Cotangent Function}} }} {{eqn | r = \frac {\frac e 2 + \frac 1 {2 e} } {\left({\frac e 2 - \frac 1 {2 e} }\right) i}       | c = Cosine of $i$ and Sine of $i$ }} {{eqn | r = \left({\frac {e + \frac 1 e} {e - \frac 1 e} }\right) \left({\frac 1 i}\right)       | c = multiplying denominator and numerator by $2$ }} {{eqn | r = \left({ \frac {e^2 + 1} {e^2 - 1} }\right) \left({\frac 1 i}\right)       | c = multiplying denominator and numerator by $e$ }} {{eqn | r = \left({ \frac {1 + e^2} {1 - e^2} }\right) i       | c = Reciprocal of $i$ }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \cos^2 x + \sin^2 x       | r = 1       | c = Sum of Squares of Sine and Cosine }} {{eqn | ll= \leadsto       | l = \sin^2 x       | r = 1 - \cos^2 x }} {{eqn | ll= \leadsto       | l = \sin x       | r = \pm \sqrt {1 - \cos^2 x} }} {{end-eqn}} Then from Sign of Sine: {{begin-eqn}} {{eqn | l = \sin x       | o = >       | r = 0       | c = if there exists an integer $n$ such that $2 n \pi < x < \paren {2 n + 1} \pi$ }} {{eqn | l = \sin x       | o = <       | r = 0       | c = if there exists an integer $n$ such that $\paren {2 n + 1} \pi < x < \paren {2 n + 2} \pi$ }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \sec^2 x - \tan^2 x       | r = 1       | c = Difference of Squares of Secant and Tangent }} {{eqn | ll= \leadsto       | l = \sec^2 x       | r = \tan^2 x + 1 }} {{eqn | ll= \leadsto       | l = \sec x       | r = \pm \sqrt {\tan ^2 x + 1} }} {{end-eqn}} 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. {{qed}}
For the first part, if there exists an integer $n$ such that $\paren {2 n - \dfrac 1 2} \pi < x < \paren {2 n + \dfrac 1 2} \pi$: {{begin-eqn}} {{eqn | l = \cos x       | r = +\frac 1 {\sqrt {1 + \tan^2 x} }       | c = Cosine in terms of Tangent }} {{eqn | ll= \leadsto       | l = \frac 1 {\paren {\frac 1 {\cos x} } }       | r = +\frac 1 {\sqrt {1 + \tan^2 x} } }} {{eqn | ll= \leadsto       | l = \frac {\sin x} {\paren {\frac {\sin x} {\cos x} } }       | r = +\frac 1 {\sqrt {1 + \tan^2 x} }       | c = multiplying denominator and numerator by $\sin x$ }} {{eqn | ll= \leadsto       | l = \frac {\sin x} {\tan x}       | r = + \frac 1 {\sqrt {1 + \tan^2 x} }       | c = Tangent is Sine divided by Cosine }} {{eqn | ll= \leadsto       | l = \sin x       | r = + \frac {\tan x} {\sqrt {1 + \tan^2 x} } }} {{end-eqn}} For the second part, if there exists an integer $n$ such that $\paren {2 n + \dfrac 1 2} \pi < x < \paren {2 n + \dfrac 3 2} \pi$: {{begin-eqn}} {{eqn | l = \cos x       | r = -\frac 1 {\sqrt {1 + \tan^2 x} }       | c = Cosine in terms of Tangent }} {{eqn | ll= \leadsto       | l = \frac 1 {\paren {\frac 1 {\cos x} } }       | r = -\frac 1 {\sqrt {1 + \tan^2 x} } }} {{eqn | ll= \leadsto       | l = \frac {\sin x} {\paren {\frac {\sin x} {\cos x} } }       | r = -\frac 1 {\sqrt {1 + \tan^2 x} }       | c = multiplying denominator and numerator by $\sin x$ }} {{eqn | ll= \leadsto       | l = \frac {\sin x} {\tan x}       | r = -\frac 1 {\sqrt {1 + \tan^2 x} }       | c = Tangent is Sine divided by Cosine }} {{eqn | ll= \leadsto       | l = \sin x       | r = -\frac {\tan x} {\sqrt {1 + \tan^2 x} } }} {{end-eqn}} When $\cos x = 0$, $\tan x$ is undefined. {{qed}}
{{begin-eqn}} {{eqn | l = \frac 1 {\sin \theta}       | r = \csc \theta       | c = Cosecant is Reciprocal of Sine }} {{eqn | ll= \leadsto       | l = \sin \theta       | r = \frac 1 {\csc \theta} }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \frac 1 {\tan \theta}       | r = \cot \theta       | c = Cotangent is Reciprocal of Tangent }} {{eqn | ll= \implies       | l = \tan \theta       | r = \frac 1 {\cot \theta} }} {{end-eqn}} $\tan \theta$ is not defined when $\cos \theta = 0$, and $\cot \theta$ is not defined when $\sin \theta = 0$. {{qed}}
{{begin-eqn}} {{eqn | l = \frac 1 {\cos \theta}       | r = \sec \theta       | c = Secant is Reciprocal of Cosine }} {{eqn | ll= \leadsto       | l = \cos \theta       | r = \frac 1 {\sec \theta} }} {{end-eqn}} $\sec \theta$ and $\dfrac 1 {\cos \theta}$ are not defined when $\cos \theta = 0$. {{qed}}
{{begin-eqn}} {{eqn | l = \sin \paren {a + b i}       | r = \sin a \cos \paren {b i} + \cos a \sin \paren {b i}       | c = Sine of Sum }} {{eqn | r = \sin a \cosh b + \cos a \sin \paren {b i}       | c = Hyperbolic Cosine in terms of Cosine }} {{eqn | r = \sin a \cosh b + i \cos a \sinh b       | c = Hyperbolic Sine in terms of Sine }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \sin a \cosh b + i \cos a \sinh b       | r = \frac {e^{i a} - e^{-i a} } {2 i} \frac {e^b - e^{-b} } 2 + i \frac {e^{i a} + e^{-i a} } 2 \frac {e^b - e^{-b} } 2       | c = Sine Exponential Formulation, Cosine Exponential Formulation, Definition of Hyperbolic Sine, Definition of Hyperbolic Cosine }} {{eqn | r = \frac {e^{b + i a} - e^{-b + i a} - e^{b - i a} + e^{-b - i a} - e^{b + i a} + e^{-b + i a} - e^{b - i a} + e^{-b - i a} } {4 i}       | c = simplifying }} {{eqn | r = \frac {e^{-b - i a} - e^{b - i a} } {2 i}       | c = simplifying }} {{eqn | r = \frac {e^{i \left({a + b i}\right)} - e^{-i \left({a + b i}\right)} } {2 i}       | c =  }} {{eqn | r = \sin \left({a + b i}\right)       | c = Sine Exponential Formulation }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \sinh \paren {a + b i}       | r = \sinh a \cosh \paren {b i} + \cosh a \sinh \paren {b i}       | c = Hyperbolic Sine of Sum }} {{eqn | r = \sinh a \cos b + \cosh a \sin \paren {b i}       | c = Cosine in terms of Hyperbolic Cosine }} {{eqn | r = \sinh a \cos b + i \cosh a \sin b       | c = Sine in terms of Hyperbolic Sine }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \cos \paren {a + b i}       | r = \cos a \cos \paren {b i} - \sin a \sin \paren {b i}       | c = Cosine of Sum }} {{eqn | r = \cos a \cosh b - \sin a \sin \paren {b i}       | c = Hyperbolic Cosine in terms of Cosine }} {{eqn | r = \cos a \cosh b - i \sin a \sinh b       | c = Hyperbolic Sine in terms of Sine }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \cos a \cosh b - i \sin a \sinh b       | r = \frac {e^{i a} + e^{-i a} } 2 \frac {e^b + e^{-b} } 2 - i \frac {e^{i a} - e^{-i a} } {2 i} \frac {e^b - e^{-b} } 2       | c = Cosine Exponential Formulation, {{Defof|Hyperbolic Cosine}}, Sine Exponential Formulation, {{Defof|Hyperbolic Sine}} }} {{eqn | r = \frac {e^{b + i a} + e^{-b + i a} + e^{b - i a} + e^{-b - i a} - e^{b + i a} + e^{-b + i a} + e^{b - i a} - e^{-b - i a} } 4       | c = simplifying }} {{eqn | r = \frac {e^{-b + i a} + e^{b - i a} } 2       | c = simplifying }} {{eqn | r = \frac {e^{i \paren {a + b i} } + e^{-i \paren {a + b i} } } 2       | c =  }} {{eqn | r = \cos \paren {a + b i}       | c = Cosine Exponential Formulation }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \map \cosh {a + b i}       | r = \cosh a \map \cosh {b i} + \sinh a \map \sinh {b i}       | c = Hyperbolic Cosine of Sum }} {{eqn | r = \cosh a \cos b + \sinh a \map \sinh {b i}       | c = Cosine in terms of Hyperbolic Cosine }} {{eqn | r = \cosh a \cos b + i \sinh a \sin b       | c = Sine in terms of Hyperbolic Sine }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \cos^2 x + \sin^2 x       | r = 1       | c = Sum of Squares of Sine and Cosine }} {{eqn | ll= \leadsto       | l = \cos^2 x       | r = 1 - \sin^2 x }} {{eqn | ll= \leadsto       | l = \cos x       | r = \pm \sqrt {1 - \sin^2 x} }} {{end-eqn}} Then from Sign of Cosine: {{begin-eqn}} {{eqn | l = \cos x       | o = >       | r = 0       | c = if there exists an integer $n$ such that $\paren {2 n - \dfrac 1 2} \pi < x < \paren {2 n + \dfrac 1 2} \pi$ }} {{eqn | l = \cos x       | o = <       | r = 0       | c = if there exists an integer $n$ such that $\paren {2 n + \dfrac 1 2} \pi < x < \paren {2 n + \dfrac 3 2} \pi$ }} {{end-eqn}} {{qed}}
For the first part, if there exists integer $n$ such that $n \pi < x < \paren {n + \dfrac 1 2} \pi$: {{begin-eqn}} {{eqn | l = \tan x       | r = +\sqrt {\sec^2 x - 1}       | c = Tangent in terms of Secant }} {{eqn | ll= \leadsto       | l = \frac {\sin x} {\cos x}       | r = +\sqrt {\sec^2 x - 1}       | c = Tangent is Sine divided by Cosine }} {{eqn | ll= \leadsto       | l = \sin x \sec x       | r = +\sqrt {\sec^2 x - 1}       | c = Secant is Reciprocal of Cosine }} {{eqn | ll= \leadsto       | l = \sin x       | r = +\frac {\sqrt {\sec^2 x - 1} } {\sec x} }} {{end-eqn}} For the second part, if there exists integer $n$ such that $\paren {n + \dfrac 1 2} \pi < x < \paren {n + 1} \pi$: {{begin-eqn}} {{eqn | l = \tan x       | r = -\sqrt {\sec^2 x - 1}       | c = Tangent in terms of Secant }} {{eqn | ll= \leadsto       | l = \frac {\sin x} {\cos x}       | r = -\sqrt {\sec^2 x - 1}       | c = Tangent is Sine divided by Cosine }} {{eqn | ll= \leadsto       | l = \sin x \sec x       | r = -\sqrt {\sec^2 x - 1}       | c = Secant is Reciprocal of Cosine }} {{eqn | ll= \leadsto       | l = \sin x       | r = -\frac {\sqrt {\sec^2 x - 1} } {\sec x} }} {{end-eqn}} When $\cos x = 0$, $\sec x$ is undefined. {{qed}}
For each $n \ge n_0$, let $\map {P'} n$ be defined as: :$\map {P'} n := \map P {n_0} \land \dots \land \map P n$ It suffices to show that $\map {P'} n$ is true for all $n \ge n_0$. It is immediate from the assumption $\map P {n_0}$ that $\map {P'} {n_0}$ is true. Now suppose that $\map {P'} n$ holds. By $(2)$, this implies that $\map P {n + 1}$ holds as well. Consequently, $\map {P'} n \land \map P {n + 1} = \map {P'} {n + 1}$ holds. Thus by the Principle of Mathematical Induction: :$\map {P'} n$ holds for all $n \ge n_0$ as desired. {{Qed}}
Define $T$ as: :$T = \set {n \in \Z : \forall k: n_0 \le k \le n: k \in S}$ Since $n \le n$, it follows that $T \subseteq S$. Therefore, it will suffice to show that: :$\forall n \ge n_0: n \in T$ Firstly, we have that $n_0 \in T$ {{iff}} the following condition holds: :$\forall k: n_0 \le k \le n_0 \implies k \in S$ Since $n_0 \in S$, it thus follows that $n_0 \in T$. Now suppose that $n \in T$; that is: :$\forall k: n_0 \le k \le n \implies k \in S$ By $(2)$, this implies: :$n + 1 \in S$ Thus, we have: :$\forall k: n_0 \le k \le n + 1 \implies k \in S$ {{MissingLinks|Closed Interval of Naturally Ordered Semigroup with Successor equals Union with Successor for $\Z$}} Therefore, $n + 1 \in T$. Hence, by the Principle of Finite Induction: :$\forall n \ge n_0: n \in T$ as desired. {{Qed}}
From Tangent is Sine divided by Cosine: :$\tan x = \dfrac {\sin x} {\cos x}$ Since $n$ is an integer, $n$ is either odd or even. === Case 1 === Let $n$ be odd. Hence let $m$ be an integer such that $n = 2 m + 1$. ==== Case 1.1 ==== :$n \pi < x < \paren {n + \dfrac 1 2} \pi \implies \paren {2 m + 1} \pi < x < \paren {2 m + \dfrac 3 2} \pi$ From Sign of Sine, $\sin x$ is negative. From Sign of Cosine, $\cos x$ is negative. Therefore: :$\tan x = \dfrac {\sin x} {\cos x} > 0$ ==== Case 1.2 ==== :$\paren {n + \dfrac 1 2} \pi < x < \paren {n + 1} \pi \implies \paren {2 m + \dfrac 3 2} \pi < x < \paren {2 m + 2} \pi$ From Sign of Sine, $\sin x$ is negative. From Sign of Cosine, $\cos x$ is positive. Therefore: :$\tan x = \dfrac {\sin x} {\cos x} < 0$ === Case 2 === Let $n$ be even. Hence let $m$ be an integer such that $n = 2 m$. ==== Case 2.1 ==== :$n \pi < x < \paren {n + \dfrac 1 2} \pi \implies 2 m \pi < x < \paren {2 m + \dfrac 1 2} \pi$ From Sign of Sine, $\sin x$ is positive. From Sign of Cosine, $\cos x$ is positive. Therefore: :$\tan x = \dfrac {\sin x} {\cos x} > 0$ ==== Case 2.2 ==== :$\paren {n + \dfrac 1 2} \pi < x < \paren {n + 1} \pi \implies \paren {2 m + \dfrac 1 2} \pi < x < \paren {2 m + 1} \pi$ From Sign of Sine, $\sin x$ is positive. From Sign of Cosine, $\cos x$ is negative. Therefore: :$\tan x = \dfrac {\sin x} {\cos x} < 0$ Therefore: {{begin-eqn}} {{eqn | l = \tan x       | o = >       | r = 0       | c = if there exists integer $n$ such that $n \pi < x < \paren {n + \dfrac 1 2} \pi$ }} {{eqn | l = \tan x       | o = <       | r = 0       | c = if there exists integer $n$ such that $\paren {n + \dfrac 1 2} \pi < x < \paren {n + 1} \pi$ }} {{end-eqn}} {{qed}}
Let $x < 0$. {{AimForCont}} $\dfrac 1 x > 0$. Then: {{begin-eqn}} {{eqn | l = x       | o = <       | r = 0       | c =  }} {{eqn | ll= \leadsto       | l = x \times \dfrac 1 x       | o = <       | r = 0 \times 0       | c = Real Number Ordering is Compatible with Multiplication: Negative Factor }} {{eqn | ll= \leadsto       | l = 1       | o = <       | r = 0       | c = Real Number Axioms: $\R \text M 4$: Inverse }} {{end-eqn}} But from Real Zero is Less than Real One: :$1 > 0$ Therefore by Proof by Contradiction: :$\dfrac 1 x < 0$ {{qed}}