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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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sai-kartik
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underdisplayname
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    You are correct that $\sqrt{x^2} \neq x$ otherwise $1 = \sqrt{(-1)^2} =-1$. Indeed, $\sqrt{x^2}=|x|$. However, if $x\in [-\pi/2,\pi/2]$ then Symbolab and your professor are correct as . – Surb Jul 22 '20 at 07:10
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    Symbolab is not correct as I did not specify the domain. It might be a glitch that they definitely need to be aware of. However, my prof was right because of the domain. Thank you so much for saving me from this irrational madness! – underdisplayname Jul 22 '20 at 07:15
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    Congrats to you for analysing such details. They are important! And they can kill your proof if you omit them. – Surb Jul 22 '20 at 07:16
  • you are right but for $t=\frac \pi 4$ we have $|\cos(t)|=\cos(t)$ – miracle173 Jul 22 '20 at 07:16

1 Answers1

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

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

Surb
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