Why is $\\sec(\\pi/2)$, when approaching from the left equal to ...
Mar 1, 2020 · and "before pi/2" means you are less than pi/2, you are on the "right" side of pi/2 considering unit circle! So, if you approach pi/2 "from right" you get infinity. Simiarly if you approach pi/2 "from left", you are on the left side of pi/2 considering unit circle and since cos function is negative there, you get minus infinity.
limits - Is the equation $\sec (\pi/2)$= $\tan (\pi/2)$ true ...
Aug 16, 2023 · While both $\sec (\pi/2)$ and $\tan (\pi/2)$ are undefined they are both basically $1/0$.In case anyone is unaware of how they equal $1/0$ what we can do is represent $\sec (\pi/2)$ as $\dfrac {1}{\cos (\pi/2)}$ and $\tan (\pi/2)$ as $\dfrac {\sin \pi/2}{\cos \pi/2}$ and once you put in the value they both become $\dfrac {1}{0}$.
Is $\\sec^{-1}(\\sec(\\pi/2)) = \\pi/2$? - Mathematics Stack Exchange
I think it shouldn't be defined as $\pi/2$ is not in the range of the function $\sec^{-1}(x)$ Wolfram confused me by giving the answer as $\pi/2$ : Link. But it mentions on another page that $\pi/2$ is not in the range of $\sec^{-1}(x)$: Link. What is the correct answer?
Why is the period of $\\sec$ and $\\csc 2\\pi$ instead of $\\pi$?
Feb 12, 2021 · When $\tan x = \frac{(2x + 1)\pi}{2}$, the function is undefined, resulting in a period of simply $\pi$. However, $\csc$ and $\sec$ also include undefined points. Taking this discontinuity into account, why are their periods $2\pi$ instead of $\pi$?
Domain of $\\sec x$ vs. domain of $\\sec^{-1}(\\sec x)$
Oct 3, 2023 · This gives us other fascinating results: for example, WolframAlpha thinks that $\sec^{-1}(1/0) = \pi/2$ as well. This is roughly equivalent to working with the projectively extended reals $\mathbb R \cup \{\tilde \infty\}$ (where the infinity in question has no sign, since -ComplexInfinity simplifies to ComplexInfinity ) and defining various ...
calculus - Is $\int_{0}^{\pi} \sec^2(x)\,dx \neq 0$? - Mathematics ...
May 8, 2018 · I have had a discussion with my classmates recently, and an answer on the 2016 Calculus BC AP test said that: $$\\int_0^\\pi \\sec^2x = \\infty$$ The problem I am having is that we can split this and w...
Evaluate : $\\int\\limits_0^{\\pi/2} \\cot x \\ln(\\sec x) dx$
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Evaluate $\\sec \\frac{2\\pi}{7}+\\sec \\frac{4\\pi}{7}+\\sec \\frac{6 ...
Jul 29, 2019 · Stack Exchange Network. Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers.
Why is the graph of $sec^2(x)/tan^2(x)$ continuous at $x=\\pi /2$?
Jun 1, 2021 · Stack Exchange Network. Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers.
Evaluate : $\\int_0^{\\frac{\\pi}{2}} \\frac{\\ln(\\sec ^2 x)\\sec^2x ...
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