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


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 $\cos(t)\geq 0$.
