$f(x)=\frac{sec^2(x)}{tan^2(x)}$
Domain of $sec^2(x)$ and $tan^2(x)$ is $\mathbb{R}-(2n+1)\frac{\pi}{2}$, for $n \in \mathbb{Z}$, hence $f(x)$ also has the same domain. I expected a discontinuity at at $x=\frac{\pi}{2}$.
But, the graph does not say so.
Graph from Wolfram Alpha:
What am I missing in my concept?
Also, even if the software did the manipulation $\frac{sec^2(x)}{tan^2(x)}$=$\frac{1}{sin^2(x)}$ it should have been valid only for $x \in \mathbb{R} -(2n+1)\frac{\pi}{2}$.
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$\begingroup$$$\frac{\sec^2(x)}{\tan^2(x)} = \frac{1}{\sin^2(x)}$$
for all values of $x$ where $\cos(x) \neq 0$. You are right in the assumption that $f(x) = \frac{\sec^2(x)}{\tan^2(x)}$ should have a discontinuity at $x = \pi/2$. Technically speaking, the function $g(x) = \frac{1}{\sin^2(x)}$ isn't technically the same function as $f$ because, as you said, they have slightly different domains. This is an instance where the computer has graphed the "simplified" function $g(x)$, exemplifying why you cannot always trust computers blindly. Your reservations are correct.
$\endgroup$ 1 $\begingroup$Yes the website is manipulating it as $\frac{\sec^2(x)}{\tan^2(x)}$=$\frac{1}{\sin^2(x)}$ but it is valid for all $ x \in \mathbb R-n\pi$ as $\sin (x) = 0 \;\;\;\forall\;\;\; x \; \in \;n\pi$ where $n \;\in\; \mathbb Z$.
But since the original function $f(x)$ itself is not defined at $x \in \mathbb R-(2n+1)\frac{\pi}{2}$ so, the graph should also be discontinuous at these values.
You are right, there may be a bug in the website's algorithm.
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