Secant in terms of Tangent
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Theorem
Let $x$ be a real number such that $\cos x \ne 0$.
Then:
| \(\ds \sec x\) | \(=\) | \(\ds +\sqrt {\tan ^2 x + 1}\) | if there exists an integer $n$ such that $\paren {2 n - \dfrac 1 2} \pi < x < \paren {2 n + \dfrac 1 2} \pi$ | |||||||||||
| \(\ds \sec x\) | \(=\) | \(\ds -\sqrt {\tan ^2 x + 1}\) | if there exists an integer $n$ such that $\paren {2 n + \dfrac 1 2} \pi < x < \paren {2 n + \dfrac 3 2} \pi$ |
where $\sec$ denotes the real secant function and $\tan$ denotes the real tangent function.
Proof
| \(\ds \sec^2 x - \tan^2 x\) | \(=\) | \(\ds 1\) | Difference of Squares of Secant and Tangent | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \sec^2 x\) | \(=\) | \(\ds \tan^2 x + 1\) | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \sec x\) | \(=\) | \(\ds \pm \sqrt {\tan ^2 x + 1}\) |
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.
$\blacksquare$