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DominiRican1013

  • one year ago

The expression (secx + tanx)2 is the same as _____.

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  1. anonymous
    • one year ago
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    \[(\sec x + \tan x)^2\] Remember how to expand out a squared expression like this? :)

  2. DominiRican1013
    • one year ago
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    \[(\frac{ 1 }{ cosx }+\frac{ sinx }{ cosx})^2 ?\]

  3. anonymous
    • one year ago
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    Before we continue, is there by chance choices?

  4. DominiRican1013
    • one year ago
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    yes hold on...

  5. anonymous
    • one year ago
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    If you're able to get past that first part, here is a hint to help you finish it up. Remember your Pythagorean identity for tangent, \[\sec ^2x=1+\tan ^2x\]

  6. DominiRican1013
    • one year ago
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    \[A. 1+2\tan^2+2secx tanx\] \[B. 1+2cscx\] \[C. \sec^2x+\tan^2x\] \[D. \sec^2x+2cscx+\tan^2x\]

  7. DominiRican1013
    • one year ago
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    These are the choices

  8. anonymous
    • one year ago
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    Well after you expand you would get \[\sec^2(x)+2\sec(x)\tan(x)+\tan^2(x)\]

  9. anonymous
    • one year ago
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    But remember after you expand everything out, you can replace the sec^2x(that you'll end up with) with 1+tan^2x.

  10. DominiRican1013
    • one year ago
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    I'm so lost What I have so far is 1+tan(x)+2sec(x)tan(x)+1

  11. anonymous
    • one year ago
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    Expand the original equation like you would a binomial

  12. anonymous
    • one year ago
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    except in this case instead of (a+b)^2 A is Sec (x) and B is Tan (x)

  13. anonymous
    • one year ago
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    look at my previous posts, they are the exact steps on how to do this.

  14. DominiRican1013
    • one year ago
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    so it's A

  15. anonymous
    • one year ago
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    yes

  16. DominiRican1013
    • one year ago
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    thank you

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