Is this equation correct for all values of x?

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The equation tan^{-1}|tan x| = |x| is not correct for all permissible values of x. The discussion highlights that tangent is not a one-to-one function, leading to discrepancies in values when using the inverse tangent. Specifically, examples show that the absolute value of the tangent function does not equate to the tangent of the absolute value of x. Additionally, the range of the atan function limits the validity of the equation, particularly for values of |x| outside [0, pi/2). Therefore, the equation fails to hold true universally.
utkarshakash
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Homework Statement


Is this equation correct for permissible values of x

tan^{-1}|tan x| = |x|

Homework Equations



The Attempt at a Solution


I assume LHS to be θ.
Then tanθ=|tanx|
The original equation becomes
tan^{-1}tan \theta = |x|
 
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hi utkarshakash! :smile:

(what do you mean by "permissible"? :confused:)

wouldn't it be easier to start by saying |tanx| = tan|x| ? :wink:
 
No, that is not correct because tangent is not a "one-to-one function". For example, if \theta= 5\pi/4 then tan(\theta)= tan(5\pi/4)= 1 so that tan^{-1}(tan(\theta))= tan^{-1}(tan(5\pi/4))= tan^{-1}(1)= \pi/4, not 5\pi/4. Since everything is positive, the absolute value is irrelevant.
 
tiny-tim said:
hi utkarshakash! :smile:

(what do you mean by "permissible"? :confused:)

wouldn't it be easier to start by saying |tanx| = tan|x| ? :wink:


Even if so, it is still wrong, as tangent can be negative for positive values of x. For instance, \tan(7\pi/4)=-1. It is easy to observe from here that |\tan(7\pi/4)|\neq \tan|7\pi/4|, as 1\neq-1.
 
Thanks all for your answers.
 
Another way to look at it is that atan(|y|) always produces a result in the range [0,pi/2). So the statement must be false for any |x| outside that range.
 
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