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Math Help - What identity should I use?

  1. #1
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    What identity should I use?

    What Identity should I use to solve this.

    csc x-\frac{cotx}{secx} =sinx
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  2. #2
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    Quote Originally Posted by mtt0216 View Post
    What Identity should I use to solve this.

    csc x-\frac{cotx}{secx} =sinx
    You are not solving anything. You are proving an identity. There's a big difference in meaning between these two words.

    I suggest you substitute the definitions of csc x, cot x and sec x into the left hand side and simplify the result.
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  3. #3
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    I meant to say prove sorry about that.

    The problem I have with this problem is that I don't know how to change

    \frac{cotx}{secx} into a different form.

    Its just the fact thats its in an equation and I just don't know where to start.
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  4. #4
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    Quote Originally Posted by mtt0216 View Post
    I meant to say prove sorry about that.

    The problem I have with this problem is that I don't know how to change

    \frac{cotx}{secx} into a different form.

    Its just the fact thats its in an equation and I just don't know where to start.
    Do you know the definition of each in terms of sinx and cosx? It will be in your notes and/or textbook. Then make the substitutions as I said in my first reply.
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  5. #5
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    I know the definitions:

    csc x = 1/sinx

    sec x = 1/cosx

    Cot= cosx/sinx

    But once I put them into the equation I dont know what to do.
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  6. #6
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    Quote Originally Posted by mtt0216 View Post
    I know the definitions:

    csc x = 1/sinx

    sec x = 1/cosx

    Cot= cosx/sinx

    But once I put them into the equation I dont know what to do.
    At this level you should be able to do basic algebra on expressions like (a/b)/(c/d) to get (ad)/(bc).

    Note: \frac{a/b}{c/d} = \frac{a}{b} \div \frac{c}{d} = \frac{a}{b} \times \frac{d}{c} = ....
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  7. #7
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    Quote Originally Posted by mr fantastic View Post
    At this level you should be able to do basic algebra on expressions like (a/b)/(c/d) to get (ad)/(bc).

    Note: \frac{a/b}{c/d} = \frac{a}{b} \div \frac{c}{d} = \frac{a}{b} \times \frac{d}{c} = ....
    OK sorrryy that I forgot a basic step.......................
    Last edited by mtt0216; March 8th 2010 at 11:44 AM.
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  8. #8
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    Quote Originally Posted by mtt0216 View Post
    I know the definitions:

    csc x = 1/sinx

    sec x = 1/cosx

    Cot= cosx/sinx

    But once I put them into the equation I dont know what to do.
    as already mentioned...

     <br />
\frac{1}{\sin{\theta}} <br />
-\frac<br />
{\frac{\cos{\theta}}{\sin{\theta}}}<br />
{\frac{1}{\cos{\theta}}}<br />
\rightarrow<br />
\frac{1}{\sin{\theta}}-\frac{\cos{\theta}}{\sin{\theta}}<br />
\frac{\cos{\theta}}{1}<br />
\rightarrow<br />
\frac{1-\cos^2{\theta}}{\sin{\theta}}<br />
\rightarrow<br />
\frac{\sin^2{\theta}}{\sin{\theta}}= \sin{\theta}<br />

    EDITprobably already got this
    Last edited by bigwave; March 8th 2010 at 11:50 AM. Reason: late reply
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  9. #9
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    Yea thats what I had. thanks
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