# Trig Identity

• Aug 17th 2010, 07:13 PM
danksnaks
Trig Identity
Tried all I could on this one. I think I'm making it out to be harder than it's supposed to be...

(sec(x))/(sec(x)-tan(x))=sec^2(x)+sec(x)tan(x)

Any help would be greatly appreciated.
• Aug 17th 2010, 08:08 PM
Gusbob
Hey, try expressing sec(x) and tan(x) in terms of sines and cosines first, cancel what you can.

You should get $\displaystyle \frac{1}{1 - \sin (x)}$

The trick here is to multiply the numerator and denominator by $\displaystyle 1 + \sin (x)$

Then use the identity $\displaystyle \sin^2 x + \cos^2 x = 1$ on the denominator.

Convert back to sec and tan and that should be it.
• Aug 17th 2010, 08:09 PM
Prove It
$\displaystyle \sec^2{x} + \sec{x}\tan{x} = \sec{x}(\sec{x} + \tan{x})$

$\displaystyle = \frac{\sec{x}(\sec{x} + \tan{x})(\sec{x} - \tan{x})}{\sec{x} - \tan{x}}$

$\displaystyle = \frac{\sec{x}(\sec^2{x} -\tan^2{x}) }{\sec{x} - \tan{x}}$

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

$\displaystyle = \frac{\sec{x}}{\sec{x} - \tan{x}}$.
• Aug 18th 2010, 05:55 AM
danksnaks
Thanks a lot guys. Your help was a godsend.