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Math Help - Identities

  1. #1
    Member purplec16's Avatar
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    Identities

    \frac{cos \beta}{1- sin\beta}= sec \beta + tan \beta
    \frac {1}{sec\beta} / 1- \frac {1}{csc\beta}
    \frac {1}{1-sec\beta}  \frac{csc\beta}{csc\beta -1}
    Is this right so far?...what do I do next?
    Last edited by purplec16; March 7th 2010 at 04:14 PM. Reason: latex problems
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  2. #2
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    \frac{\cos{b}}{1-\sin{b}} \cdot \frac{1+\sin{b}}{1+\sin{b}} =

    \frac{\cos{b}(1 + \sin{b})}{1 - \sin^2{b}} =

    \frac{\cos{b}(1 + \sin{b})}{\cos^2{b}} =

    \frac{1 + \sin{b}}{\cos{b}} =

    \frac{1}{\cos{b}} + \frac{\sin{b}}{\cos{b}}
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  3. #3
    Member purplec16's Avatar
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    Thanks so much, how do u you understand how to do these things, when you do it I undestand, but when the question is there on its own I dont know what to do, I guess it calls for a lot of practice
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  4. #4
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    Quote Originally Posted by purplec16 View Post
    Thanks so much, how do u you understand how to do these things, when you do it I undestand, but when the question is there on its own I dont know what to do, I guess it calls for a lot of practice
    bingo!
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  5. #5
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    Hi purplec16,

    you can start to reason it through a little also...

    \frac{cos\beta}{1-sin\beta}=sec\beta+tan\beta

    We can write out the definitions for the terms on the right hand side...

    \frac{cos\beta}{1-sin\beta}=\frac{1}{cos\beta}+\frac{sin\beta}{cos\b  eta}

    Now we see we need to have cos\beta under the line on the left hand side,

    therefore as we can only multiply by 1 (so the ratio stays the same),

    we multiply the left hand side by \frac{cos\beta}{cos\beta}

    as this gets us that term in the denominator,
    however we see later on, that this term comes in anyway, so it ends up being unnecessary!

    \frac{cos\beta}{1-sin\beta}\ \frac{cos\beta}{cos\beta}=\frac{cos^2\beta}{cos\be  ta(1-sin\beta)}

    Now we are not supposed to have 1-sin\beta under the line
    so we need to fish that out!

    We therefore multiply by 1 again, this time using the "conjugate" of 1-sin\beta

    because (1-sin\beta)(1+sin\beta)=1(1+sin\beta)-sin\beta(1+sin\beta)=1+sin\beta-sin\beta-sin^2\beta

    =1-sin^2\beta which is cos^2\beta

    Therefore

    \frac{cos^2\beta}{cos\beta(1-sin\beta)}\ \frac{1+sin\beta}{1+sin\beta}=\frac{cos^2\beta(1+s  in\beta)}{cos^3\beta}

    =\frac{1+sin\beta}{cos\beta}=\frac{1}{cos\beta}+\f  rac{sin\beta}{cos\beta}
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