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Math Help - Verifying Identities

  1. #1
    Sabs
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    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.

    Thanks in advance!!
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Sabs View Post
    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
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  3. #3
    Math Engineering Student
    Krizalid's Avatar
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    Another way.

    Quote Originally Posted by Sabs View Post

    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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