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Simplifying a trigonometry expression

  1. Aug 25, 2013 #1
    1. The problem statement, all variables and given/known data

    Simplify:

    [(sec x)(sec2 x) - (tan x - 1)(sec x tan x)] /
    sec2 x


    2. Relevant equations

    none

    3. The attempt at a solution

    [(sec x)(sec2 x) - (tan x - 1)(sec x tan x)] /
    sec2 x

    [(sec x)(sec2 x) - tan x (1 -1) sec x ] /
    sec2 x

    [(sec x)(sec2 x) - 0 ] /
    sec2 x

    sec x

    Thanks! I hope the above is correct, but it probably isn't.
     
  2. jcsd
  3. Aug 25, 2013 #2

    haruspex

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    That step is wrong. You've dropped a tan x. Try it again.
    But I would start by getting rid of the denominator. What is another way of expressing 1/sec(θ)?
     
  4. Aug 25, 2013 #3

    HallsofIvy

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    So
    [tex]\frac{sec(x)(sec^2(x))- (tan(x)- 1)(sec(x)tan(x))}{sec^2(x)}[/tex]?

    Is there any particular reason for writing "[itex]sec(x)sec^2(x)[/itex]" rather than [itex]sec^3(x)[/itex]?

    Where did that "(1- 1)" come from?
    sec2 x

    [(sec x)(sec2 x) - 0 ] /
    sec2 x

    sec x

    Thanks! I hope the above is correct, but it probably isn't.[/QUOTE]
    It's easy to check that it is NOT correct. If x= 30 degrees ([itex]\pi/6[/itex] radians), the original expression is [itex](2/3)\sqrt{3}+ 2/3[/itex] while sec(x) is [itex](2/3)\sqrt{3}[/itex].
     
  5. Aug 25, 2013 #4

    HallsofIvy

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    So
    [tex]\frac{sec(x)(sec^2(x))- (tan(x)- 1)(sec(x)tan(x))}{sec^2(x)}[/tex]?

    Is there any particular reason for writing "[itex]sec(x)sec^2(x)[/itex]" rather than [itex]sec^3(x)[/itex]?

    Where did that "(1- 1)" come from?

    It's easy to check that it is NOT correct. If x= 30 degrees ([itex]\pi/6[/itex] radians), the original expression is [itex](2/3)\sqrt{3}+ 2/3[/itex] while sec(x) is [itex](2/3)\sqrt{3}[/itex].
     
  6. Aug 26, 2013 #5
    Thanks haruspex, and sorry for my late reply. I have a hard time getting online in the evenings.

    cos x = 1/sec x

    So:

    [(sec x)(sec2 x) - (tan x - 1)(sec x tan x)] /
    sec2 x

    [(sec x)(sec2 x) - (tan x - 1)(sec x tan x)] /
    [(sec x)(sec x)]

    (cos x)(cos x) [(sec x)(sec2 x) - (tan x - 1)(sec x tan x)]

    cos2 x [sec3 x - (sec2 x tan x - sec x tan x)]

    [sec3 x/sec2 x] - [[(sec x)(tan2 x - tan x)]/sec2 x]

    sec x - [[(tan2 x - tan x)]/sec x]

    sec x - [cos x[(tan2 x - tan x)]]

    That's it. I don't know how to reduce this more.
     
    Last edited: Aug 26, 2013
  7. Aug 26, 2013 #6
    maybe you can simplify it this way:

    [itex]1-\left(\frac{\tan^2(x)-\tan(x)}{1+\tan^2(x)}\right)[/itex]

    [itex]\frac{1+\tan^2(x)-\tan^2(x)+\tan(x)}{1+\tan^2(x)}=\frac{1+\tan(x)}{1+\tan^2(x)}[/itex]
     
  8. Aug 26, 2013 #7
    Thanks for your reply HallsofIvy!

    That is the way the problem was presented to me on the homework!

    I tried to distribute the tan x out of both factors!
     
  9. Aug 26, 2013 #8

    haruspex

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    There's a mistake in how you multiplied out the (tan x - 1)(sec x tan x) term.
    Why flip back to dividing by sec x again? You can use sec x * cos x = 1 without having to turn it back into a division.
     
  10. Aug 26, 2013 #9
    Ok, I decided to redo the problem. Part of my problem is it's hard to check the work on here because I can't write fractions they way they are supposed to be written without complex computer operations.

    Thank you harupex for the tip: sec x * cos x = 1

    I knew they were inverses, but it didn't dawn on me to use that equation!

    [(sec x)(sec2 x) - (tan x - 1)(sec x tan x)]
    /
    sec2 x

    cos2 x [(sec3 x) - (tan x - 1) (sec x tan x)]

    sec x - cos2 x (sec x tan2 x - sec x tan x)

    sec x - (cos x tan2 x + cos x tan x)

    sec x - [(cos x tan x)(tan x + 1)]
     
  11. Aug 26, 2013 #10
    Thanks janhaa! I am not sure what you did though to get rid of the initial "1 -".
     
  12. Aug 26, 2013 #11

    haruspex

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    There's a sign error in that step.
    What does cos x tan x simplify to?
    Do you know sec2 = 1 + tan2 ? (Can derive this easily from cos2+sin2=1.)
     
  13. Aug 26, 2013 #12
    Ok, I see my error. I become very flustered with trig for some reason. I guess I get anxious when the pressure is on.

    Ok, since tan x = sin x/cos x,
    cos x tan x reduces to just sin x.

    Here it is corrected:

    sec x - cos2 x (sec x tan2 x - sec x tan x)

    sec x - cos x tan2 x + cos x tan x

    sec x - cos x tan2 x + sin x

    And yes, I know that sec2 = 1 + tan2 is one of the three Pythagorean identities, but I don't see any sec2 or 1's above.
     
  14. Aug 26, 2013 #13

    haruspex

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    No, but there's a tan2.
     
  15. Aug 26, 2013 #14
    Oh my gosh. I think I have it!

    sec x - cos x tan2 x + sin x

    sec x - cos x (sec2 x -1) + sin x

    sec x - cos x sec2 x - cos x + sin x

    sec x -sec x - cos x + sin x

    sin x - cos x


    I hope it's that easy.

    For some reason, I am not seeing all the basic trigonometric equivalents.
     
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