# Is this equation correct

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## Homework Statement

Is this equation correct for permissible values of x

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

## The Attempt at a Solution

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

## Answers and Replies

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tiny-tim
Science Advisor
Homework Helper
hi utkarshakash!

(what do you mean by "permissible"? )

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

HallsofIvy
Science Advisor
Homework Helper
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.

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 $1\neq-1$.

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Thanks all for your answers.

haruspex
Science Advisor
Homework Helper
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2020 Award
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.