Sign of Tangent

Theorem
Let $x$ be a real number.

Then:

where $\tan$ denotes the tangent function.

Proof
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
If $n$ is odd, let $m$ be an integer such that $n = 2 m + 1$.

Case 1.1

 * $n \pi < x < \left({n + \dfrac 1 2}\right) \pi \implies \left({2 m + 1}\right) \pi < x < \left({2 m + \dfrac 3 2}\right) \pi$

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

 * $\left({n + \dfrac 1 2}\right) \pi < x < \left({n + 1}\right) \pi \implies \left({2 m + \dfrac 3 2}\right) \pi < x < \left({2 m + 2}\right) \pi$

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
If $n$ is even, let $m$ be an integer such that $n = 2 m$.

Case 2.1

 * $n \pi < x < \left({n + \dfrac 1 2}\right) \pi \implies 2 m \pi < x < \left({2 m + \dfrac 1 2}\right) \pi$

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

 * $\left({n + \dfrac 1 2}\right) \pi < x < \left({n + 1}\right) \pi \implies \left({2 m + \dfrac 1 2}\right) \pi < x < \left({2 m + 1}\right) \pi$

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: