1. ## identities

prove the ff identities

sec x + 1 / tan x - tan x / sec x -1 = 0

1-sinx/1+sinx = (secx-tanx)^2

thanks

2. Hi,
Originally Posted by eepyej
prove the ff identities

sec x + 1 / tan x - tan x / sec x -1 = 0
$\displaystyle \frac{\sec(x)+1}{\tan(x)}-\frac{\tan(x)}{\sec(x)-1}=\frac{(\sec(x)+1) (\sec(x)-1)-\tan^2(x)}{\tan(x)(\sec(x)-1)}$

Now, expand in the numerator $\displaystyle (\sec(x)+1)(\sec(x)-1)$ thanks to this formula : $\displaystyle (a+b)(a-b)=a^2-b^2$
And remember that $\displaystyle 1+\tan^2(x)=\sec^2(x)$

1-sinx/1+sinx = (secx-tanx)^2
Multiply by $\displaystyle 1=\frac{1-\sin(x)}{1-\sin(x)}$

3. Hello, eepyej!

Another approach . . .

Prove: .$\displaystyle \frac{\sec x + 1}{\tan x} - \frac{\tan x}{\sec x -1} \;=\; 0$

Multiply the first fraction by $\displaystyle \frac{\sec x - 1}{\sec x - 1}$

. . $\displaystyle {\color{blue}\frac{\sec x - 1}{\sec x - 1}}\cdot\frac{\sec x + 1}{\tan x} - \frac{\tan x}{\sec x - 1} \;\;=\;\;\frac{\sec^2\!x-1}{\tan x(\sec x-1)} - \frac{\tan x}{\sec x - 1}$

. . $\displaystyle = \;\frac{\tan^2\!x}{\tan x(\sec x - 1)} - \frac{\tan x}{\sec x - 1} \;\;=\;\;\frac{\tan x}{\sec x - 1} - \frac{\tan x}{\sec x - 1} \;\;=\;\;0$