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Trigonometric Identity Problems

  1. Jan 4, 2007 #1
    Can anyone help me solve the following problems?

    sec theta -1/1-cos theta = sec theta

    tan (pie/2 - theta) tan theta =1

    Thanks
     
  2. jcsd
  3. Jan 5, 2007 #2

    James R

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    Your first equation is ambiguous. Could you please add appropriate brackets so we know where the fraction is?

    For the second one, do you know the identity for tan(A+B)?

    Because that would be a good place to start.
     
  4. Jan 5, 2007 #3

    dextercioby

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    For the second, use the definition of the tangent function.

    Daniel.
     
  5. Jan 5, 2007 #4

    HallsofIvy

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    Robert R's point is that it is not clear whether you mean sec theta -1/(1-cos theta) = sec theta or (sec theta -1)/(1- cos theta). What is the DEFINITION of sec theta? Then try multiplying both sides of the equation by cos theta.
     
  6. Jan 5, 2007 #5
    Krystal i think you meant:

    for question 1

    [sec(theta) - 1] / [1 - cos(theta)] = sec(theta)

    if you did then:
    try writing sec(theta) as 1/cos(theta)
    then make a common denominator for: 1/cos(theta) -1
    and continue from there
     
  7. Jul 16, 2007 #6
    I have a problem that I cannot get past step one; the problem is verifying the identity:

    cos2x(1+cot^2x) = csc^2 - 2 ?????

    I know that [1 + cot^2x] = [csc^2x]

    Thanks for all the help.
     
  8. Jul 16, 2007 #7
    nevermind...i got the answer...it was:

    (1-2sin^2 x)/sin^2 x = (1-2sin^2 x)/sin^2 x
     
  9. Jul 25, 2011 #8
    can anyone help me to solve this problem?
    (tanѲ/1-cotѲ)+(cotѲ/1-tanѲ)=1+secѲcscѲ
     
  10. Jul 25, 2011 #9

    SammyS

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    You have posted this on a 4 year old thread.

    You may get more help, if you start a new thread and post it there.
     
  11. Jul 26, 2011 #10
    trigonometric solver

    (tanѲ/1-cotѲ)+(cotѲ/1-tanѲ)=1+secѲcscѲ
     
  12. Jul 26, 2011 #11

    SammyS

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    Without the proper placement of parentheses, or other grouping symbols, what you have written is equivalent to: [itex]\displaystyle \left(\frac{\tan\theta}{1}-\cot\theta\right)+\left(\frac{\cot\theta}{1}-\tan\theta\right)=1+\sec\theta\,\csc\theta\,.[/itex]

    Perhaps you meant: (tanθ/{1-cotθ})+(cotθ/(1-tanθ))=1+secθcscθ, which is the same as: [itex]\displaystyle \left(\frac{\tan\theta}{1-\cot\theta}\right)+\left(\frac{\cot\theta}{1-\tan\theta}\right)=1+\sec\theta\,\csc\theta\,.[/itex]
     
  13. Jul 26, 2011 #12

    SammyS

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    To show that [itex]\displaystyle \left(\frac{\tan\theta}{1-\cot\theta}\right)+\left(\frac{\cot\theta}{1-\tan\theta}\right)=1+\sec\theta\,\csc\theta\,[/itex] is an identity, I suggest writing the tangent & cotangent functions in terms of sine & cosine.
     
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