# How do you verify the identity (tanx - secx + 1)/(tanx + secx - 1) = cosx/(1 + sinx)?

##### 1 Answer
Apr 30, 2018

$L H S = \frac{\tan x - \sec x + 1}{\tan x + \sec x - 1}$

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

$= \frac{\tan x - \sec x + 1}{\tan x + \sec x - \left(\sec x - \tan x\right) \left(\sec x + \tan x\right)}$

=(cancel(tanx - secx + 1))/((tanx + secx)(cancel(1 - secx+tanx))

$= \frac{1}{\sec x + \tan x}$

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

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

$= \cos \frac{x}{1 + \sin x} = R H S$

Alternative method

$L H S = \frac{\tan x - \sec x + 1}{\tan x + \sec x - 1}$

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

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

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

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

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

$= \cos \frac{x}{1 + \sin x} = R H S$