Shape of Tangent Function

Theorem
The nature of the tangent function on the set of real numbers $$\mathbb{R}$$ is as follows:


 * $$\tan x$$ is strictly increasing on the interval $$\left[{-\frac \pi 2 \, . \, . \, \frac \pi 2}\right]$$;
 * $$\tan x$$ is not defined on $$\forall n \in \mathbb{Z}: x = \left({n + \frac 1 2}\right) \pi$$;
 * $$\tan x \to + \infty$$ as $$x \to \frac \pi 2 ^-$$;
 * $$\tan x \to - \infty$$ as $$x \to -\frac \pi 2 ^+$$.

Proof

 * $$\tan x$$ is strictly increasing on $$\left[{-\frac \pi 2 \, . \, . \, \frac \pi 2}\right]$$:

This has been demonstrated in the discussion on Tangent Function is Periodic on Reals.


 * $$\tan x$$ is not defined on $$x = \left({n + \frac 1 2}\right) \pi$$:

From the discussion of Sine and Cosine are Periodic on Reals, it was established that $$\forall n \in \mathbb{Z}: x = \left({n + \frac 1 2}\right) \pi \Longrightarrow \cos x = 0$$.

As division by zero is not defined, it follows that at these points $$\tan x$$ is not defined either.


 * $$\tan x \to + \infty$$ as $$x \to \frac \pi 2 ^-$$:

From Sine and Cosine are Periodic on Reals, we have that both $$\sin x > 0$$ and $$\cos x > 0$$ on $$\left({0 \, . \, . \, \frac \pi 2}\right)$$.

We have that:
 * 1) $$\cos x \to 0$$ as $$x \to \frac \pi 2 ^-$$;
 * 2) $$\sin x \to 1$$ as $$x \to \frac \pi 2 ^-$$.

Thus it follows that $$\tan x = \frac {\sin x} {\cos x} \to + \infty$$ as $$x \to \frac \pi 2 ^-$$.


 * $$\tan x \to - \infty$$ as $$x \to -\frac \pi 2 ^+$$:

From Sine and Cosine are Periodic on Reals, we have that $$\sin x < 0$$ and $$\cos x > 0$$ on $$\left({-\frac \pi 2 \, . \, . \, 0}\right)$$.

We have that:
 * 1) $$\cos x \to 0$$ as $$x \to -\frac \pi 2 ^+$$;
 * 2) $$\sin x \to -1$$ as $$x \to -\frac \pi 2 ^+$$.

Thus it follows that $$\tan x = \frac {\sin x} {\cos x} \to - \infty$$ as $$x \to -\frac \pi 2 ^+$$.