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Trig problem

  1. May 9, 2006 #1

    I'm brand new here; needless to say, I think it's great what this community is doing. Now to my question: my class is just finishing up geometry, which means we're starting to enter into beginning trig. Here's the problem:

    1-Sinx/1+Sinx = (Secx-Tanx)^2

    So far, I've been able to reduce to this: Cos^2x=1

    from this point on, I'm stumped--any help would be greatly appreciated.

  2. jcsd
  3. May 9, 2006 #2


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    HINT: Multiply numerator and denominator by

    [tex]1 - \sin x[/tex]
  4. May 9, 2006 #3
    I don't understand why I would need to multiply by 1-sinx? I cleared the left side by 1+sinx => 1+Sinx(1-Sinx/1+sinx) => 1-Sin^2x which is Cos^2x--I guess what I'm trying to get at is that I'm not grasping the reason to multiply the numerator and denominator by (1-sinx)

  5. May 9, 2006 #4


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    Try it and see what happens. Also, after you have done this, glance through some trig idents and see what jumps out.

  6. May 9, 2006 #5


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    Uhmm, I don't understand what you wrote here... :frown:
    You are using a lot of "implies", which do not make much sense in this case.
    What do you mean by:
    [tex]1 + \sin x \Rightarrow (1 + \sin x) \frac{1 - \sin x}{1 + \sin x} \Rightarrow 1 - \sin ^ 2 x \Rightarrow \cos ^ 2 x[/tex]??? :confused:
    Have you tried cancelling (1 - sin x) from both sides?
    [tex]\frac{1 - sin x}{1 + \sin x} = (\sec x - \tan x) ^ 2[/tex]
    [tex]\Leftrightarrow \frac{1 - sin x}{1 + \sin x} = \left( \frac{1}{\cos x} - \frac{\sin x}{\cos x} \right) ^ 2[/tex]
    [tex]\Leftrightarrow \frac{1 - sin x}{1 + \sin x} = \frac{(1 - \sin x) ^ 2}{\cos ^ 2 x}[/tex]
    You should be able to go from here, right? Can you? :)
  7. May 9, 2006 #6


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    The first thing you should make clear when given a question like this is to ascertain whether you're asked to 1) prove an identity or 2) solve an equation.

    An identity is a relationship between two expressions that is true for *all* valid x. You can't solve an identity, you can only prove that it's true. The question will be phrased "Prove that ... = ..." or "Show that ... = ..." and the goal is to manipulate either the Left Hand Side to become the Right Hand Side or vice versa (one will do).

    An equation is a relationship between two expressions that only holds true for finitely many values of the variable x. It is not true in general. The goal here is to do stuff to find out what value(s) of x the equation holds for, that's the required solution set. You cannot prove an equation, you can only solve it.

    Now, what you have looks like an identity proof question because

    [tex]\frac{1 - \sin x}{1 + \sin x} = {(\sec x - \tan x)}^2[/tex]

    holds true for all valid x.

    When you prove an identity, your proof must take the form LHS = LHS' = LHS '' = ... = RHS, where by all the prime symbols (') I mean manipulation of the Left Hand Side. It can also take the form RHS = RHS' = ... = LHS.

    Note the use of the "=" (equal) sign here, not the implied sign as VietDao pointed out.

    In contrast, when you solve an equation you should use the implied sign. If the equation is of the form LHS = RHS, then the solution will go [tex]LHS = RHS \Rightarrow LHS' = RHS' \Rightarrow ... x = (solution)[/tex]

    In each case (proving identities or solving equations), you will use various manipulations involving common trigonometric identities, which you should know in a very familiar way.

    I hope the basics are clearer now, in order to avoid confusion. :smile:
    Last edited: May 9, 2006
  8. May 9, 2006 #7
    1-sin^2x is the conjugate... the opposite of the denominator... its useful to get a difference of squares on the bottom.. if you did stuff with radicals, you probably have seen it before...
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