Is this equation correct for all values of x?

[utkarshakash]

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
$tan^{-1}tan \theta = |x|$

 

[tiny-tim]

hi utkarshakash!

(what do you mean by "permissible"? )

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

 

[HallsofIvy]

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.

 

[Millennial]

hi utkarshakash!

(what do you mean by "permissible"? )

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

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 $1\neq-1$.

 

[utkarshakash]

Thanks all for your answers.

 

[haruspex]

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.