Verifying Identities

Sabs · March 7th 2008, 05:45 PM

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.

Thanks in advance!!

**topsquark** · March 7th 2008, 05:48 PM

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.

Thanks in advance!!

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

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

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

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

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

-Dan

**Krizalid** · March 8th 2008, 06:40 AM

Another way.

Originally Posted by Sabs

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

$\frac{{\sec x - 1}}<br /> {{1 - \cos x}} = \frac{{\sec x}}<br /> {{\sec x}} \cdot \frac{{\sec x - 1}}<br /> {{1 - \cos x}} = \sec x \cdot \frac{{\sec x - 1}}<br /> {{\sec x - 1}} = \sec x.$

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