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Calculation of $left(frac{1}{cos^2x}right)^{frac{1}{2}}$

Mathematics Asked by underdisplayname on November 19, 2021

Shouldn’t $left(frac{1}{cos^2x}right)^{frac{1}{2}} = |sec(x)|$?
Why does Symbolab as well as my professor (page one, also below) claim that $left(frac{1}{cos^2x}right)^{frac{1}{2}} = sec(x)$, which can be negative? Also, the length of a vector cannot be negative… isn’t it?

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One Answer

You are correct that $sqrt{x^2} neq x$ for every $xinBbb R$. Otherwise, we would have absurdity like $1 = sqrt{1}=sqrt{(-1)^2} =-1$. For $xinBbb R$ we indeed have $sqrt{x^2}=|x|$. However, we have $sqrt{x^2}=x$ for all $xgeq 0$.

Hence, as you professor seems to assume that $tin [-pi/2,pi/2]$, his claim that $sqrt{frac{1}{cos^2(x)}}=sec(t)$ is correct. Indeed for $tin [-pi/2,pi/2]$, we have $cos(t)geq 0$.

Answered by Surb on November 19, 2021

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