$\ln|\cos x| = \ln|1/\sec x| = \ln|(\sec x)^{-1}|=-\ln|\sec x|$
Is what I am doing valid? Or is it not correct because of the absolute function?
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I don't think $\log\lvert \cos x \rvert = \log \lvert \tan x \rvert$ – Edward Evans Apr 02 '16 at 13:32
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I edited the question, sorry... was suppose to be sec – user327983 Apr 02 '16 at 14:16
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No, what you are doing is wrong because $\cot x=\frac{1}{\tan x}$.
But I would like to point out that $$\ln |\cos x|=-\ln |\sec x|$$ Is true, as $|a||\frac{1}{a}|=1$ is always true. The absolute function does not hinder it one bit. (well, it does as it makes it both positive. but it does not effect the multiple).
S.C.B.
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Not at all. Suppose $\ln|\cos x|=-\ln|\cot x|$; the expression makes sense only if $\cos x\ne0$, $\cot x\ne0$ and $\cot x$ is defined, so for $x\ne k\pi/2$ (for $k$ an integer).
Then the equality is equivalent to $$ |\cos x|=\frac{1}{|\cot x|},\qquad x\ne k\pi/2 $$ that is, to $$ |\cos x|(|\sin x|-1)=0,\qquad x\ne k\pi/2 $$ No $x$ satisfies these conditions.
egreg
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