# another identity

• August 9th 2009, 03:53 PM
windir
another identity
Prove...
$\frac{1}{sec(x)-tan(x)}=sec(x)+tan(x)$

this is as far as i got
working from LHS

$\frac{1}{sec(x)-tan(x)}$

$cos(x)-cot(x)$

$cos(x)-\frac{cos(x)}{sin(x)}$

$\frac{sin(x)}{sin(x)}cos(x)-\frac{cos(x)}{sin(x)}$

$\frac{sin(x)cos(x)}{sin(x)}-\frac{cos(x)}{sin(x)}$

now i'm stuck..
• August 9th 2009, 04:05 PM
Bruno J.
You must know that $(a-b)(a+b)=a^2-b^2$. This identity is worth being very familiar with.

So you want to prove $1=\sec^2x-\tan^2x$ (I just multiplied both sides of your identity by $\sec x - \tan x$).
Now just apply the identity $\sec^2x=1+\tan^2x$ and you're done!
• August 9th 2009, 04:06 PM
Chris L T521
Quote:

Originally Posted by windir
Prove...
$\frac{1}{sec(x)-tan(x)}=sec(x)+tan(x)$

this is as far as i got
working from LHS

$\frac{1}{sec(x)-tan(x)}$

$cos(x)-cot(x)$

$cos(x)-\frac{cos(x)}{sin(x)}$

$\frac{sin(x)}{sin(x)}cos(x)-\frac{cos(x)}{sin(x)}$

$\frac{sin(x)cos(x)}{sin(x)}-\frac{cos(x)}{sin(x)}$

now i'm stuck..

$\frac{1}{\sec x-\tan x}{\color{red}\neq}\frac{1}{\sec x}-\frac{1}{\tan x}$!!!! XD

What I suggest you do with the LHS is the following: $\frac{1}{\sec x-\tan x}=\frac{1}{\sec x-\tan x}\cdot\frac{\sec x+\tan x}{\sec x+\tan x}=\dots$

In the process, invoke one of the pythagorean identities...

Can you take it from here?
• August 9th 2009, 04:08 PM
Random Variable
$\frac{1}{\sec x - \tan x} \cdot \frac{\sec x + \tan x}{\sec x + \tan x}$

$= \frac {\sec x + \tan x}{\sec^{2} x - \tan^{2} x }$

$= \frac {\sec x + \tan x}{1 + \tan^{2}x - \tan^{2}x}$

$= \sec x + \tan x$
• August 9th 2009, 04:13 PM
windir
ooooooooooohhh, duh! thanks man!
• August 11th 2009, 04:11 AM
mr fantastic
Quote:

Originally Posted by windir
Prove...
$\frac{1}{sec(x)-tan(x)}=sec(x)+tan(x)$

[snip]

Also discussed here: http://www.mathhelpforum.com/math-he...589-proof.html