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Homework Help: Trigonometric Identity

  1. May 5, 2012 #1
    1. The problem statement, all variables and given/known data

    Prove that:
    tan^2∅/tan∅ - 1 + cot^2∅/cot∅ - 1 = 1 + sec∅cosec∅

    2. Relevant equations

    3. The attempt at a solution

    I have solved the question taking tan∅ = sin∅/cos∅.
    But I want to solve it some other way.
  2. jcsd
  3. May 5, 2012 #2
    What other way??
    Why aren't you happy with your solution?
  4. May 5, 2012 #3


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    What you wrote for the left hand side is literally (tan^2∅/tan∅) - 1 + (cot^2∅/cot∅) - 1, which is equivalent to tan∅ + cot∅ - 2 .

    Assuming that you meant [itex]\displaystyle \frac{\tan^2(\phi)}{\tan(\phi)-1}+\frac{\cot^2(\phi)}{\cot(\phi)-1}=1+\sec(\phi)\csc(\phi)\ ,[/itex]

    yes there is another way. --- it's no better, but looks interesting enough. Even with it, eventually you will change tan to sin/cos or perhaps tan to sec/csc.

    Change cot(ϕ) to 1/tan(ϕ) . Then multiply the numerator & denominator of the second fraction by -tan(ϕ) --- that will give you a common denominator. You can then get a difference of cubes in the numerator ...
  5. May 5, 2012 #4
    Thanks, I got the answer. But I have got one more question:

    How to prove that slopes of perpendicular lines on graph paper have a product equal to -1 ?
  6. May 6, 2012 #5

    What is the angle(acute) between two lines of slopes say, m1 and m2?
    When will they become perpendicular then?
    Last edited: May 6, 2012
  7. May 6, 2012 #6
    No idea !!!
  8. May 6, 2012 #7
    Okay hmm, try drawing out two lines with a general angle θ between them. Say the angle the first line makes with the positive x axis is A and the second line makes an angle B, now try finding a trignometrical relation between θ, A and B. (Hint: use the property of external angles)
  9. May 6, 2012 #8
    I have attached a pic. Tell me if it is like that.

    Attached Files:

  10. May 6, 2012 #9
    The x axis is not necessarily where the two lines meet. So you can draw them cutting the x axis at different points, and such that they intersect somewhere arbitarily on the xy plane, for the sake of a more general result.
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