# How do you prove the identity (tan x + sec x - 1) / (tan x - sec x +1) = tan x + sec x?

Aug 12, 2015

${\sin}^{2} A + {\cos}^{2} A = 1$
Divide by ${\cos}^{2} A$:
${\tan}^{2} A + 1 = {\sec}^{2} A$

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