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Thread: proof

  1. #1
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    proof

    Given that $\displaystyle x = sec\theta + tan\theta $

    show that $\displaystyle \frac{1}{x} = sec\theta - tan\theta $
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  2. #2
    MHF Contributor red_dog's Avatar
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    $\displaystyle \frac{1}{x}=\frac{1}{\displaystyle\frac{1}{\cos\th eta}+\frac{\sin\theta}{\cos\theta}}=\frac{\cos\the ta}{1+\sin\theta}=$

    $\displaystyle =\frac{\cos\theta(1-\sin\theta)}{1-\sin^2\theta}=\frac{\cos\theta(1-\sin\theta)}{\cos^2\theta}=\frac{1-\sin\theta}{\cos\theta}=$

    $\displaystyle =\frac{1}{\cos\theta}-\frac{\sin\theta}{\cos\theta}=\sec\theta-\tan\theta$
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  3. #3
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    Just been thinking, is this method correct?

    $\displaystyle x = sec\theta + tan\theta $

    $\displaystyle \frac{1}{x} = sec\theta - tan\theta $

    $\displaystyle \frac{1}{sec\theta+tan\theta} = sec\theta - tan\theta $

    multiply both sides by $\displaystyle sec\theta+tan\theta $

    $\displaystyle 1 = ( sec\theta - tan\theta) (sec\theta+tan\theta ) $

    $\displaystyle \frac{1}{sec\theta - tan\theta} = x $

    $\displaystyle 1 = x(sec\theta - tan\theta) $

    $\displaystyle \frac{1}{x} = sec\theta - tan\theta $
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  4. #4
    MHF Contributor Bruno J.'s Avatar
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    This exact question was posted yesterday here : http://www.mathhelpforum.com/math-he...-identity.html
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  5. #5
    MHF Contributor Bruno J.'s Avatar
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    Quote Originally Posted by Tweety View Post
    Just been thinking, is this method correct?

    $\displaystyle x = sec\theta + tan\theta $

    $\displaystyle \frac{1}{x} = sec\theta - tan\theta $

    $\displaystyle \frac{1}{sec\theta+tan\theta} = sec\theta - tan\theta $

    multiply both sides by $\displaystyle sec\theta+tan\theta $

    $\displaystyle 1 = ( sec\theta - tan\theta) (sec\theta+tan\theta ) $

    $\displaystyle \frac{1}{sec\theta - tan\theta} = x $

    $\displaystyle 1 = x(sec\theta - tan\theta) $

    $\displaystyle \frac{1}{x} = sec\theta - tan\theta $
    You seem to be assuming that $\displaystyle
    \frac{1}{x} = \sec\theta - \tan\theta
    $ and that $\displaystyle
    {x} = \sec\theta + \tan\theta
    $, doing a few operations on these premises, undoing those operations, and then concluding that $\displaystyle
    \frac{1}{x} = \sec\theta - \tan\theta
    $.
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  6. #6
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    Quote Originally Posted by Bruno J. View Post
    This exact question was posted yesterday here : http://www.mathhelpforum.com/math-he...-identity.html

    Well than they must be using the same excercise book as me, as my exam board is with edexcel, this is the book I am using!

    Edexcel AS and A Level Modular Mathematics - Core Mathematics 3: Amazon.co.uk: Mr Keith Pledger: Books
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  7. #7
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    Quote Originally Posted by Bruno J. View Post
    You seem to be assuming that $\displaystyle
    \frac{1}{x} = \sec\theta - \tan\theta
    $ and that $\displaystyle
    {x} = \sec\theta + \tan\theta
    $, doing a few operations on these premises, undoing those operations, and then concluding that $\displaystyle
    \frac{1}{x} = \sec\theta - \tan\theta
    $.

    Well yes becasue thats what the book says?

    can you tell me what steps are not 'valid'?

    Also when I have the expression $\displaystyle 1 = ( sec\theta - tan\theta) (sec\theta+tan\theta ) $ and than divide most sides by $\displaystyle sec\theta - tan\theta $

    would that not cancel $\displaystyle sec\theta - tan\theta $ from the right hand side? so I get this expression; $\displaystyle \frac{1}{sec\theta - tan\theta} = sec\theta+tan\theta $

    thanks.
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  8. #8
    MHF Contributor Bruno J.'s Avatar
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    No, you are given that $\displaystyle
    {x} = \sec\theta + \tan\theta
    $, from which you must deduce that $\displaystyle
    \frac{1}{x} = \sec\theta - \tan\theta
    $.

    You cannot use in your proof that $\displaystyle
    \frac{1}{x} = \sec\theta - \tan\theta
    $ because that is what you are trying to show. However that is what you seem to be doing in the third line : starting from $\displaystyle
    \frac{1}{x} = \sec\theta - \tan\theta
    $, you replaced $\displaystyle x$ by $\displaystyle
    \sec\theta + \tan\theta
    $.
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  9. #9
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    Quote Originally Posted by Bruno J. View Post
    No, you are given that $\displaystyle
    {x} = \sec\theta + \tan\theta
    $, from which you must deduce that $\displaystyle
    \frac{1}{x} = \sec\theta - \tan\theta
    $.

    You cannot use in your proof that $\displaystyle
    \frac{1}{x} = \sec\theta - \tan\theta
    $ because that is what you are trying to show. However that is what you seem to be doing in the third line : starting from $\displaystyle
    \frac{1}{x} = \sec\theta - \tan\theta
    $, you replaced $\displaystyle x$ by $\displaystyle
    \sec\theta + \tan\theta
    $.

    oh I get what you meant now,

    Thanks
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