# How do you verify cos(x) / (1-sin(x)) = sec(x) + tan(x)?

Jul 26, 2018

Please see the explanation below

#### Explanation:

We need

${\sin}^{2} x + {\cos}^{2} x = 1$

$\sec x = \frac{1}{\cos} x$

$\tan x = \sin \frac{x}{\cos} x$

The $L H S = \cos \frac{x}{1 - \sin x}$

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

$= \frac{\cos x \left(1 + \sin x\right)}{1 - {\sin}^{2} x}$

$= \frac{\cos x \left(1 + \sin x\right)}{{\cos}^{2} x}$

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

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

$= \sec x + \tan x$

$= L H S$

$Q E D$

Jul 26, 2018

#### Explanation:

Here,

$L H S = \cos \frac{x}{1 - \sin x}$

$\text{Multiplynumerator and denominator by cosx}$

$L H S = \frac{\cos x \cdot \cos x}{\cos x \left(1 - \sin x\right)}$

$\textcolor{w h i t e}{L H S} = {\cos}^{2} \frac{x}{\cos x \left(1 - \sin x\right)}$

$\textcolor{w h i t e}{L H S} = \frac{1 - {\sin}^{2} x}{\cos x \left(1 - \sin x\right)} \to \left[\because {\sin}^{2} \theta + {\cos}^{2} \theta = 1\right]$

$\textcolor{w h i t e}{L H S} = \frac{\left(1 - \sin x\right) \left(1 + \sin x\right)}{\cos x \left(1 - \sin x\right)}$

$\textcolor{w h i t e}{L H S} = \frac{1 + \sin x}{\cos} x$

$\textcolor{w h i t e}{L H S} = \frac{1}{\cos} x + \sin \frac{x}{\cos} x$

$\textcolor{w h i t e}{L H S} = \sec x + \tan x$

$L H S = R H S$