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Thread: Help please

  1. #1
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    Help please

    Show that 1+tan^2x = (cosx)^-2 where defined.

    Don't have a clue what to do.
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  2. #2
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    Quote Originally Posted by jumba View Post
    Show that 1+tan^2x = (cosx)^-2 where defined.

    Don't have a clue what to do.
    To prove whether or not

    1+tan^2x=\frac{1}{cos^2x}

    then... either express tanx in terms of cosx, or vice versa

    Simplest is to express tan in terms of cos

    1+\frac{sin^2x}{cos^2x}=\frac{cos^2x}{cos^2x}+\fra  c{sin^2x}{cos^2x}

    A very important trigonometric identity completes the proof after you express that as a single fraction
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  3. #3
    Member Miss's Avatar
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    Don't make it complicated -_-
    1+tan^2(x)=sec^2(x)=\frac{1}{cos^2(x)}
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  4. #4
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    Excellent response, Miss !!

    However, \frac{1}{cosx} is "called" sec(x)

    so we must prove that

    1+tan^2(x)=sec^2(x)=\frac{1}{cos^2(x)}

    without swopping between the alternative descriptions of the same trigonometric ratio.

    You have highlighted a situation where marks could be easily lost in an exam!
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  5. #5
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    Quote Originally Posted by Miss View Post
    Don't make it complicated -_-
    1+tan^2(x)=sec^2(x)=\frac{1}{cos^2(x)}
    I don't know if you understood, jumba,

    the above is not a proof,
    it's only a statement.

    Whether you choose to call \frac{1}{cosx} as sec(x) or not, you still need to prove that it is 1+tan^2(x)
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  6. #6
    Member Miss's Avatar
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    Quote Originally Posted by Archie Meade View Post
    I don't know if you understood, jumba,

    the above is not a proof,
    it's only a statement.

    Whether you choose to call \frac{1}{cosx} as sec(x) or not, you still need to prove that it is 1+tan^2(x)
    Its an acceptable proof, not a statement !
    Its a standard identity.
    If you think like this, then you need to prove that tan^2(x)=\frac{sin^2(x)}{cos^2(x)}.
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  7. #7
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    sec(x)=\frac{1}{cos(x)}

    is a definition.

    Please don't try to confuse the student.

    Your second statement contradicts yourself.
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  8. #8
    Member Miss's Avatar
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    Quote Originally Posted by Archie Meade View Post
    sec(x)=\frac{1}{cos(x)}

    is a definition.

    Please don't try to confuse the student.

    Your second statement contradicts yourself.

    I did not confuse him
    But what I wrote is an accpetable proof.
    Am sure
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  9. #9
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    It's not a proof.

    sec(x) is the name given to \frac{1}{cos(x)}

    The question may be written

    Prove 1+tan^2(x)=\frac{1}{cos^2(x)}

    Or..

    Prove 1+tan^2(x)=sec^2(x)

    It's the same question!

    Interposing the alternative name does not outline a proof.
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  10. #10
    Member Miss's Avatar
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    Quote Originally Posted by Archie Meade View Post
    It's not a proof.

    sec(x) is the name given to \frac{1}{cos(x)}

    The question may be written

    Prove 1+tan^2(x)=\frac{1}{cos(x)}

    Or..

    Prove 1+tan^2(x)=sec(x)

    It's the same question!

    Interposing the alternative name does not outline a proof.
    Well, then \frac{sin^2(x)}{cos^2(x)} is a name given to tan^2(x). So prove that tan^2(x)=\frac{sin^2(x)}{cos^2(x)}.
    and \frac{cos(x)}{cos(x)} is a name given to 1.
    ....etc etc
    You are moving around a circle.
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  11. #11
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    Are you serious????
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  12. #12
    Member Miss's Avatar
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    Quote Originally Posted by Archie Meade View Post
    Are you serious????
    Sure.
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  13. #13
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    I reckon it's a case for a moderator then
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  14. #14
    Member Miss's Avatar
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    Quote Originally Posted by Archie Meade View Post
    I reckon it's a case for a moderator then
    OK. Do it.
    Question: Do you know how many proofs have trigonometric identities in Earth?
    According to your posts: they are not proofs, they are just statements!
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  15. #15
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    How do you show that

    sec(x)=\frac{1}{cos(x)}

    without only stating that

    sec(x)=\frac{1}{cos(x)} ?

    or without just calling it an identity.
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