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$ \sec(\sec^{-1}(x))=x $ when $x$ lies in $ (-\infty,-1] \cup [1,\infty). $

But when $x$ lies in $(-1,0),$ what will this function return?

My textbook says it becomes $-x$, but I don't understand how.

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    Dont you mean $\sec^{-1}(x)$? – 2017-02-24
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    i would say the textbook is wrong. (-1,1) does not belong to the domain of the function – 2017-02-24
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    I edited your question but had to make some changes. Please check that what I did is what you meant. – 2017-02-24
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    @Arbuja Yes, my bad. Getting used to MathJax still. – 2017-02-24
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    @B.Goddard Thanks for that – 2017-02-24
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    @AmritanshSinghal Can you tell how to get the x inside the domain? For example, in problems related to arcsin, we can add or subtract mutiples of $ pi $ to bring the value inside the domain, right? – 2017-02-24
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    @Shashank actually, if (-1,1) doesn't belong to the domain, then there exists no value of f(x) for that set. Notice carefully, your question says sec(arcsecx) and not arcsec(secx). – 2017-02-24

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You are right

$\sec(\sec^{-1}(x))=x$ when $x\in(-\infty,1]\bigcup[1,\infty)$

This is the domain of the function, where the points are defined. Any points where $x$ lies between $(-1,1)$ is undefined.

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    I think you misinterpret the question. (-1,1) means all real numbers between -1 and -1. And there exists no value of arcsecx when x=-0.5 for example – 2017-02-24