# (tan x+sec x -1)/(tan x -sec x +1 ) =tan x+ sec x ?

##### 2 Answers
May 23, 2018

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

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

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

$= \tan x + \sec x = R H S$

May 23, 2018

In the explanation section.

#### Explanation:

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

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

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

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

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

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

=$\frac{2 \tan x \cdot \left(\sec x + \tan x\right)}{2 \tan x}$

=$\sec x + \tan x$