1. ## Verifying Identities

I've been verifying Identities of trig equations all night! And there's but ONE that I can't get!!!!

sec(x)-1 = sec(x)
1-cos(x)

I'm supposed to make the left look like the right, basically. Can anyone solve this for me, with an explanation?? It's really got me stumped.

2. Originally Posted by Sabs
I've been verifying Identities of trig equations all night! And there's but ONE that I can't get!!!!

sec(x)-1 = sec(x)
1-cos(x)

I'm supposed to make the left look like the right, basically. Can anyone solve this for me, with an explanation?? It's really got me stumped.

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

$= \frac{\frac{1}{cos(x)} - 1}{1 - cos(x)}$

$= \frac{\frac{1}{cos(x)} - 1}{1 - cos(x)} \cdot \frac{cos(x)}{cos(x)}$

$= \frac{1 - cos(x)}{(1 - cos(x))~cos(x)}$

$= \frac{1}{cos(x)} = sec(x)$
so long as $cos(x) \neq 1$

-Dan

3. Another way.

Originally Posted by Sabs

sec(x)-1 = sec(x)
1-cos(x)
$\frac{{\sec x - 1}}
{{1 - \cos x}} = \frac{{\sec x}}
{{\sec x}} \cdot \frac{{\sec x - 1}}
{{1 - \cos x}} = \sec x \cdot \frac{{\sec x - 1}}
{{\sec x - 1}} = \sec x.$