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    Is it because of the modulus?
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    It is indeed because of the modulus. It makes the sign meaningless.
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    Yes.
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    Indeed. In fact, you don't even need the modulus signs there any more, because the top and bottom are both square (and thus positive, assuming they're not complex), so you have \dfrac{1}{2} \ln \left[ \dfrac{(\sin x + 1)^2}{\cos^2 x} \right]

    EDIT: This can in fact be further simplified to give just \ln \left| \dfrac{\sin x + 1}{\cos x} \right|, since a\ln b = \ln b^a (but you have to reintroduce the mod as you want the positive square root).
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    NOTE THAT |\frac{x}{y}|=\frac{|x|}{|y|}

    DO YOU GET ME?
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    (Original post by nuodai)
    Indeed. In fact, you don't even need the modulus signs there any more, because the top and bottom are both square (and thus positive, assuming they're not complex), so you have \dfrac{1}{2} \ln \left[ \dfrac{(\sin x + 1)^2}{\cos^2 x} \right]

    EDIT: This can in fact be further simplified to give just \ln \left| \dfrac{\sin x + 1}{\cos x} \right|, since a\ln b = \ln b^a (but you have to reintroduce the mod as you want the positive square root).
    I disagree with your reasoning. It is possible to have -cos^2x

    Your reasoning would be correct if it had been (-cos(x).(-cos(x))
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    (Original post by steve2005)
    I disagree with your reasoning. It is possible to have -cos^2x

    Your reasoning would be correct if it had been (-cos(x).(-cos(x))
    What's wrong with his reasoning? Isn't he essentially saying that  |\pm m^2| = m^2 , which is true (for real m)?
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    (Original post by GHOSH-5)
    What's wrong with his reasoning? Isn't he essentially saying that  |\pm m^2| = m^2 , which is true (for real m)?
    He first says "In fact, you don't even need the modulus signs there any more"

    With the modulus of course the expression is positive BUT nuodai argument was the squares can't be negative and of course they can be negative.
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    (Original post by steve2005)
    With the modulus of course the expression is positive BUT nuodai argument was the squares can't be negative and of course they can be negative.
    Firstly, squares of real numbers can't be negative. Secondly, he got rid of the minus sign, because he was working from the expression on the RHS of the OP's post.
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    (Original post by generalebriety)
    Firstly, squares of real numbers can't be negative. Secondly, he got rid of the minus sign, because he was working from the expression on the RHS of the OP's post.
    I think you should read the following, rather, than jump to incorrect conclusions about what I was saying.

    "Indeed. In fact, you don't even need the modulus signs there any more, because the top and bottom are both square (and thus positive, assuming they're not complex), so you have "
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    (Original post by steve2005)
    I think you should read the following, rather, than jump to incorrect conclusions about what I was saying.

    "Indeed. In fact, you don't even need the modulus signs there any more, because the top and bottom are both square (and thus positive, assuming they're not complex), so you have "
    So, like I said, he was very clearly talking about the RHS of the OP's equality.
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    (Original post by mrmanps)

    Is it because of the modulus?
    generalebriety, I am well aware of your eminence but sometimes you do jump to conclusions that are not justified.

    Please, before you defend an error, make sure you understand the point of dispute.
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    (Original post by steve2005)
    generalebriety, I am well aware of your eminence but sometimes you do jump to conclusions that are not justified.

    Please, before you defend an error, make sure you understand the point of dispute.
    So what exactly are you saying?

    It's certainly true that  \displaystyle \left | \dfrac{(\sin x+1)^2}{\cos^2 x} \right | = \dfrac{(\sin x+1)^2}{\cos^2 x}

    which is what nuodai has said (for real x).
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    (Original post by steve2005)
    generalebriety, I am well aware of your eminence but sometimes you do jump to conclusions that are not justified.

    Please, before you defend an error, make sure you understand the point of dispute.
    I don't understand the point you're making. There are only two potential conclusions to be drawn from nuodai's post: one is that he agreed with the OP, and then gave a further simplification while referring to the fraction where the top and bottom were both squares; the other is that he believed -cos^2(x) to be everywhere non-negative, and then dropped a minus sign when writing his post, combining one stupid mistake and one careless mistake to accidentally obtain the right answer. I know nuodai to be a competent mathematician, and so prefer the former interpretation. What about you?
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    The modulus make it positive.

    However, "Originally Posted by nuodai
    Indeed. In fact, you don't even need the modulus signs there any more, because the top and bottom are both square (and thus positive, assuming they're not complex), so you "

    I agree top and bottom are squares, however, denominator could be negative. It is perfectly possible to have -cos^2(x)

    In other words the expression is positive because of the modulus sign and not because it involves squares.
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    (Original post by steve2005)
    The modulus make it positive.

    However, "Originally Posted by nuodai
    Indeed. In fact, you don't even need the modulus signs there any more
    He is still very clearly talking about the RHS, not the LHS. And on the RHS, it is the squares that make it positive.
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    (Original post by generalebriety)
    He is still very clearly talking about the RHS, not the LHS. And on the RHS, it is the squares that make it positive.
    I'm sorry that I have failed to explain why I disagree with the reasoning given. It is of course true that if you square a negative number then the outcome is positive, however, it is perfectly possible to have  -x^2

    The expression is positive because of the modulus sign , end of.
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    (Original post by steve2005)
    I'm sorry that I have failed to explain why I disagree with the reasoning given. It is of course true that if you square a negative number then the outcome is positive, however, it is perfectly possible to have  -x^2
    Yes, but the OP doesn't have any minus signs on the RHS of his original equality. You're really failing to grasp this, aren't you?
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    (Original post by mrmanps)

    Is it because of the modulus?
    My answer to the OP is that the negative disappears because of the modulus sign, and that is the only reason.

    All discussions about squares are irrelevant.
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    (Original post by steve2005)
    I agree top and bottom are squares, however, denominator could be negative. It is perfectly possible to have -cos^2(x)
    Since when has -cos^2 x been a square? (given we've said we're not considering complex numbers).

    In other words the expression is positive because of the modulus sign and not because it involves squares.
    Which expression? \left|\frac{(1+\sin x)^2}{-\cos^2 x}\right| or \left|\frac{(1+\sin x)^2}{\cos^2 x}\right|.

    Given nuodai's ability, it is unreasonable to suggest he was talking about the former.
 
 
 
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