Proof of The Tangent: Derivative, Monotonicity, Bijectivity onto $\mathbb{R}$ and Special Values

Routine arithmetic and routine facts about functions are used without comment (Routine Facts, Continued: Typed Arithmetic, Maps and Predicativity); results stated for Derivative of a Real Function on an Interval are read for Derivative of a Real Function on an Interval by The Earlier and the New Versions of the Definitions of the Derivative on an Interval and of the Riemann Integral Agree (ii) and The Earlier and the New Versions of the Definitions of the Derivative on an Interval and of the Riemann Integral Agree (iii).

Write $I=(-\pi/2,\pi/2)$. Since $\pi>0$ by Quarter-Turn Identities and Periodicity of Sine and Cosine (i), $I$ is an interval containing the two distinct points $0$ and $\pi/4$, so derivatives of maps $I\to\mathbb{R}$ are defined. Moreover

$$\cos x>0\qquad(x\in I)$$

by The Cosine Is Positive on $(-\pi/2,\pi/2)$ (i); this is what makes $\tan x=\sin x/\cos x$ meaningful on $I$, and it will be used repeatedly below.

The derivative. By Continuity, Derivatives and Power Series of Sine and Cosine (ii), $\sin$ and $\cos$ are differentiable on $\mathbb{R}$, with $\sin'=\cos$ and $\cos'=-\sin$. Their restrictions $\sin|_{I}$ and $\cos|_{I}$ to the subinterval $I$ are therefore differentiable at every $x\in I$, relative to $I$, with the same derivatives $\cos x$ and $-\sin x$, by Restriction Stability of Continuity and of the Derivative (ii). On $I$ the tangent is the quotient $\sin|_{I}/\cos|_{I}$, and its denominator has no zero on $I$. The quotient rule Derivatives of Constants, the Identity, Sums, Scalar Multiples, Products, Reciprocals, Quotients and Powers (viii), applied with numerator $\sin|_{I}$ and denominator $\cos|_{I}$, shows that $\tan$ is differentiable at every $x\in I$, with

$$\tan'(x)=\frac{\cos x\cdot\cos x-\sin x\cdot(-\sin x)}{\cos^{2}x}=\frac{\cos^{2}x+\sin^{2}x}{\cos^{2}x}.$$

The numerator equals $1$ by the Pythagorean identity The Pythagorean Identity for Sine and Cosine (i), which gives $\tan'(x)=1/\cos^{2}x$; splitting the fraction instead gives

$$\tan'(x)=1+\frac{\sin^{2}x}{\cos^{2}x}=1+\tan^{2}x .$$

Thus $\tan$ is differentiable on $I$, and its derivative takes the values claimed in (i).

Continuity and strict increase. Being differentiable at every point of $I$, $\tan$ is continuous at every point of $I$ by The Increment Form of the Derivative, and Differentiability Implies Continuity (ii), that is, continuous on $I$. Its derivative $1/\cos^{2}x$ is positive at every $x\in I$. Hence The Sign and Size of the Derivative: Monotonicity, Constancy and Lipschitz Bounds (ii), applied to the continuous map $\tan$ on the interval $I$, gives $\tan x<\tan y$ for all $x,y\in I$ with $x<y$: $\tan$ is strictly increasing on $I$. This proves (ii).

Unbounded above. We show that for every $M>0$ there is $x\in I$ with $\tan x>M$. Near $\pi/2$ the denominator $\cos x$ is small and positive while the numerator $\sin x$ is close to $1$, and we make this precise. The functions $\cos$ and $\sin$ are continuous at $\pi/2$, by Continuity, Derivatives and Power Series of Sine and Cosine (i), and

$$\cos\frac{\pi}{2}=0,\qquad \sin\frac{\pi}{2}=1$$

by Quarter-Turn Identities and Periodicity of Sine and Cosine (ii). Let $M>0$ be given. Choose first, by continuity of $\cos$ at $\pi/2$, a $\delta_{1}>0$ with $|\cos t|<1/(2M)$ for every $t\in\mathbb{R}$ with $|t-\pi/2|<\delta_{1}$; then choose, by continuity of $\sin$ at $\pi/2$, a $\delta_{2}>0$ with $|\sin t-1|<1/2$, and so $\sin t>1/2$, for every $t\in\mathbb{R}$ with $|t-\pi/2|<\delta_{2}$; and put $\delta=\min\{\delta_{1},\delta_{2},\pi/2\}$. The point $x=\pi/2-\delta/2$ satisfies $\pi/4\le x<\pi/2$, so it lies in $(0,\pi/2)\subseteq I$, and $|x-\pi/2|=\delta/2$ is less than both $\delta_{1}$ and $\delta_{2}$. Since $x\in I$, the cosine is positive there, so $0<\cos x<1/(2M)$ and $\sin x>1/2$, and therefore

$$\tan x=\frac{\sin x}{\cos x}>\frac{1/2}{1/(2M)}=M .$$

Unbounded below. The interval $I$ is symmetric about $0$: for $x\in I$ also $-x\in I$. By Values at Zero, Parity and Double-Angle Identities for Sine and Cosine (ii), $\sin(-x)=-\sin x$ and $\cos(-x)=\cos x$, so

$$\tan(-x)=\frac{-\sin x}{\cos x}=-\tan x\qquad(x\in I).$$

Given $M>0$, take $x\in I$ with $\tan x>M$ as in the previous paragraph; then $-x\in I$ and $\tan(-x)<-M$.

Bijectivity. Being strictly increasing, $\tan$ is injective: if $x\neq y$ in $I$, say $x<y$, then $\tan x<\tan y$. Since $\tan$ is continuous on the interval $I$, its image $J=\tan(I)$ is an interval, by Intermediate Value Theorem on a Closed Real Interval, and Continuous Images of Intervals (ii). Now let $y\in\mathbb{R}$ and put $M=|y|+1>0$. The two previous paragraphs give points $x_{1},x_{2}\in I$ with

$$\tan x_{1}<-M<y<M<\tan x_{2},$$

so $y$ lies between two elements of $J$, and $y\in J$ because $J$ is an interval. Hence $\tan$ maps $I$ onto $\mathbb{R}$; being also injective, it is a bijection of $I$ onto $\mathbb{R}$, which is (iii).

Special values. By Values at Zero, Parity and Double-Angle Identities for Sine and Cosine (i), $\sin 0=0$ and $\cos 0=1$, so $\tan 0=0/1=0$. For $\pi/4$, apply the quarter-turn identity Quarter-Turn Identities and Periodicity of Sine and Cosine (iii) at the point $-\pi/4$, and then Values at Zero, Parity and Double-Angle Identities for Sine and Cosine (ii):

$$\sin\frac{\pi}{4}=\sin\Bigl(-\frac{\pi}{4}+\frac{\pi}{2}\Bigr)=\cos\Bigl(-\frac{\pi}{4}\Bigr)=\cos\frac{\pi}{4}.$$

Since $\pi/4\in I$, $\cos(\pi/4)>0$, so we may divide and obtain $\tan(\pi/4)=1$. This proves (iv).