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    Given that x = sec ϴ+ tan ϴ show that x + 1/x = 2 sec ϴ
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    (Original post by Hirsty97)
    Given that x = sec ϴ+ tan ϴ show that x + 1/x = 2 sec ϴ
    What have you tried? Please post your working.
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    I'd start by writing \displaystyle x = \sec\theta + \tan\theta as a single fraction. Remember that \displaystyle \sec\theta = \frac 1 {\cos\theta} and \displaystyle \tan\theta = \frac {\sin\theta}{\cos\theta}. This'll make evaluating \displaystyle x + \frac 1 x slightly nicer.
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    (Original post by Notnek)
    What have you tried? Please post your working.
    I don't know. I get sec ϴ + tan ϴ + 1/sec ϴ + tan ϴ. I figure I have to get it all as a single fraction
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    1/X is 1/sec(theta)+tan(theta). Multiply top and bottom by sectheta-tan theta and you will get sectheta-tantheta over sec^2theta-tan^2theta. And we know from the identity of tan^2+1=sec^2 that sec^2-tan^2=1. You simply substitute it in simplify and get your answer.
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    (Original post by Hirsty97)
    Given that x = sec ϴ+ tan ϴ show that x + 1/x = 2 sec ϴ
    Try expressing (secϴ + tanϴ) as a single fraction. You know that secϴ = 1/cosϴand that tanϴ= sinϴ/cosϴ. Both of these have the same denominator, so you can group them together. To find 1/x, all you have to do is turn the whole fraction upside down.
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    (Original post by Radioactivedecay)
    1/X is 1/sec(theta)+tan(theta). Multiply top and bottom by sectheta-tan theta and you will get sectheta-tantheta over sec^2theta-tan^2theta. And we know from the identity of tan^2+1=sec^2 that sec^2-tan^2=1. You simply substitute it in simplify and get your answer.
    Textbook says first step is:

    (Sec ϴ + tan ϴ)^2 + 1 / sec ϴ + tan ϴ

    I get what to do then but the answer is apparently 2 sec ϴ
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    (Original post by Hirsty97)
    Textbook says first step is:

    (Sec ϴ + tan ϴ)^2 + 1 / sec ϴ + tan ϴ

    I get what to do then but the answer is apparently 2 sec ϴ
    What did you do from there?
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    (Original post by _gcx)
    What did you do from there?
    Expand the brackets

    Then sec^2 ϴ = tan^2ϴ + 1

    So you get sec^2ϴ + 2secϴtanϴ + sec^2ϴ \ secϴ + tanϴ

    Which you can factories and cancel out to be left with 2 sec ϴ

    I get that just not the first part to go from:
    (Sec ϴ + tan ϴ) + 1/ sec ϴ + tan ϴ

    To:
    (Sec ϴ + tan ϴ) ^2 + 1/ sec ϴ + tan ϴ
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    (Original post by Hirsty97)
    Expand the brackets

    Then sec^2 ϴ = tan^2ϴ + 1

    So you get sec^2ϴ + 2secϴtanϴ + sec^2ϴ \ secϴ + tanϴ

    Which you can factories and cancel out to be left with 2 sec ϴ

    I get that just not the first part to go from:
    Sec ϴ + tan ϴ + 1/ sec ϴ + tan ϴ

    To:
    (Sec ϴ + tan ϴ) ^2 + 1/ sec ϴ + tan ϴ
    \displaystyle (\sec\theta + \tan\theta) + \frac 1 {\sec\theta + \tan\theta} = \frac{(\sec\theta + \tan\theta)^2}{\sec\theta + \tan\theta}  + \frac 1 {\sec\theta + \tan\theta} = \frac{(\sec\theta + \tan\theta)^2 + 1}{\sec\theta + \tan\theta}
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    (Original post by _gcx)
    \displaystyle (\sec\theta + \tan\theta) + \frac 1 {\sec\theta + \tan\theta} = \frac{(\sec\theta + \tan\theta)^2}{\sec\theta + \tan\theta}  + \frac 1 {\sec\theta + \tan\theta} = \frac{(\sec\theta + \tan\theta)^2 + 1}{\sec\theta + \tan\theta}
    Yeah that's it. So I just multiply it by the denominator?
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    (Original post by Hirsty97)
    Yeah that's it. So I just multiply it by the denominator?
    I wouldn't multiply by anything at this stage.

    Expand the square bracket on the numerator and use a standard trig formula that relates tan^2 to sec^2.
 
 
 
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