# How do you solve sec(x)-1=tan(x)?

Mar 6, 2018

$x = 2 k \pi , k \in Z$

#### Explanation:

$\sec x - 1 = \tan x \Rightarrow \frac{1}{\cos} x - 1 = \sin \frac{x}{\cos} x \Rightarrow 1 - \cos x = \sin x \Rightarrow \cos x + \sin x = 1 \Rightarrow \frac{1}{\sqrt{2}} \cos x + \frac{1}{\sqrt{2}} \sin x = \frac{1}{\sqrt{2}}$
$\Rightarrow \cos x \cos \left(\frac{\pi}{4}\right) + \sin x \sin \left(\frac{\pi}{4}\right) = \cos \left(\frac{\pi}{4}\right)$
$\Rightarrow \cos \left(x - \frac{\pi}{4}\right) = \cos \left(\frac{\pi}{4}\right)$
$x - \frac{\pi}{4} = 2 k \pi \pm \frac{\pi}{4} , k \in Z$
$x = 2 k \pi \pm \frac{\pi}{4} + \frac{\pi}{4} , k \in Z$
$x = 2 k \pi + \frac{\pi}{4} + \frac{\pi}{4} , k \in Z \mathmr{and} x = 2 k \pi - \frac{\pi}{4} + \frac{\pi}{4.} k \in Z$
$\textcolor{red}{x = 2 k \pi + \frac{\pi}{2} , k \in Z \mathmr{and} x = 2 k \pi , k \in Z}$
But, taking $k = 0 , . . e t c .$
$x = 2 k \pi + \frac{\pi}{2.} k \in Z \implies x = \frac{\pi}{2}$, which does not satisfy
$\sec x - 1 = \tan x$, as $\sec \left(\frac{\pi}{2}\right) \mathmr{and} \tan \left(\frac{\pi}{2}\right)$ are undefined.
So, $\textcolor{red}{x \ne 2 k \pi + \frac{\pi}{2} , k \in Z \Rightarrow x = 2 k \pi , k \in Z}$

Mar 6, 2018

$x = 2 k \pi , \text{where k any integer}$

#### Explanation:

sec(x)−1=tan(x)

1/cosx −1=sinx/cosx

$\frac{1 - \sin x}{\cos} x = 1$

sinx + cosx = 1, "where" cosx ≠ 0

$x = 2 k \pi , \text{where k any integer}$