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Prove the Identity (cosecA+cotA)^2 similar to 1+cosA/1-cosA

  1. Dec 11, 2009 #1
    1. The problem statement, all variables and given/known data

    Hi all , again i am stuck onto this question :( , tried over 3 sheets alone on it lol.btw. thanks for your replies ;) .
    Prove the Identity (cosecA+cotA)^2 similar to 1+cosA/1-cosA
    2. Relevant equations
    hmm lets see.. sin2+cos2=1 ,
    sec2= 1+tan2
    cosec2= 1+cot2, cot=1/tanx= cosx/sinx


    3. The attempt at a solution

    i started off my expanding brackets:
    => cosec2A+cot2A+2cosecAcotA
    and the rest well =/ no idea.. i couldn't prove them, just had a thought while typing cant i like sub . some value before expanding them ? ooh =/
     
  2. jcsd
  3. Dec 11, 2009 #2

    rock.freak667

    User Avatar
    Homework Helper

    Re: Trigonometry

    Write everything in terms of sinA and cosA, then simplify some more.
     
  4. Dec 11, 2009 #3

    Mark44

    Staff: Mentor

    Re: Trigonometry

    Do you mean "=" where you have written "similar to?"
    Also, you need parentheses on the right side. What you have would be correctly interpreted as 1 + (cosA/1) - cosA, but I'm pretty sure that's not what you intended.
    At least give some hint that you are talking about the squares of functions. Without any context sin2 would be interpreted as the sine of 2 radians.
    You started with (cscA + cotA)^2. This is equal to csc^2(A) + 2cscAcotA + cot^2(A). When you write cosec2A, most people would read this as cosec(2A) (usually written as csc(2A)).
     
  5. Dec 11, 2009 #4
    Re: Trigonometry

    Ah. again thanks a zillion :)
    i think i got it..:
    (cosecA + cotA) ^2
    (1/SinA + CosA/SinA )^2
    =>(1+cosA^2 /sin^2A) <-- i am doubtful about this bit
    => 1+cos2A/1-cos2A
     
  6. Dec 11, 2009 #5
    Re: Trigonometry

    oh sorry about that, next time i will use latex.
     
  7. Dec 11, 2009 #6

    Mark44

    Staff: Mentor

    Re: Trigonometry

    Here's what you want to say
    (cosecA + cotA) ^2 = (1/SinA + CosA/SinA )^2
    = [(1 + cosA)/sinA]^2
    When you square 1 + cosA you will have three terms. What you have below has only two terms.
     
  8. Dec 11, 2009 #7
    Re: Trigonometry

    yea (a+b)^2 formula so am i on the right track so far?
     
  9. Dec 11, 2009 #8

    Mark44

    Staff: Mentor

    Re: Trigonometry

    Up to that point, yes.
     
  10. Dec 11, 2009 #9
    Re: Trigonometry

    this is what i got:
    [(1+cosA / Sin A )]^2
    => 1+ cos^2A + 2cosA / Sin^2 A
    => hmm would i take cos as common?
     
  11. Dec 11, 2009 #10

    Mark44

    Staff: Mentor

    Re: Trigonometry

    You are using ==> ("implies") when you should be using =.

    You have
    [tex]\frac{(1 + cosA)^2}{sin^2A}[/tex]
    Leave the numerator in factored form (don't expand it).
    Rewrite the denominator using an identity that you know.
    Factor the denominator.
    Simplify.
     
  12. Dec 11, 2009 #11
    Re: Trigonometry

    AH... thanks alot , i get it now( much clearer ).
    Solved .
     
  13. Dec 11, 2009 #12

    Mark44

    Staff: Mentor

    Re: Trigonometry

    Your work that you turn in should look pretty much like this:
    [tex](cscA + cotA)^2~=~(\frac{1}{sinA}~+~\frac{cosA}{sinA})^2~=~\frac{(1 + cosA)^2}{sin^2A}~=~\frac{(1 + cosA)^2}{(1 - cosA)^2}~=~\frac{(1 + cosA)(1 + cosA)}{(1 - cosA)(1 + cosA)}~=~\frac{1 + cosA}{1 - cosA}[/tex]

    To see my LaTeX code, click what I have above and a new window will open that has my script.
     
  14. Dec 12, 2009 #13
    Re: Trigonometry


    Roger that , to be honest i cant thank you enough =), anyhow i will be back again ( thats for sure using LATEX code) =).
     
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