Tangent in terms of Secant
Jump to navigation
Jump to search
Theorem
Let $x$ be a real number such that $\cos x \ne 0$.
Then:
| \(\ds \tan x\) | \(=\) | \(\ds +\sqrt {\sec^2 x - 1}\) | if there exists an integer $n$ such that $n \pi < x < \paren {n + \dfrac 1 2} \pi$ | |||||||||||
| \(\ds \tan x\) | \(=\) | \(\ds -\sqrt {\sec^2 x - 1}\) | if there exists an integer $n$ such that $\paren {n + \dfrac 1 2} \pi < x < \paren {n + 1} \pi$ |
where $\tan$ denotes the real tangent function and $\sec$ denotes the real secant function.
Proof
| \(\ds \sec^2 x - \tan^2 x\) | \(=\) | \(\ds 1\) | Difference of Squares of Secant and Tangent | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \tan^2 x\) | \(=\) | \(\ds \sec^2 x - 1\) | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \tan x\) | \(=\) | \(\ds \pm \sqrt {\sec^2 x - 1}\) |
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.
$\blacksquare$