Answer
The function is discontinuous at the following points:
$x = \pi~n,~~$ where $n$ is an integer
$x = \frac{\pi}{2}+\pi~n,~~$ where $n$ is an integer
Work Step by Step
$y = ln(tan^2~x)$
Note that $ln~x$ is continuous on the interval $(0, \infty)$ and $ln~x$ is undefined on the interval $(-\infty, 0]$
The function $y$ is discontinuous when $tan^2~x = 0$ or $tan^2~x$ is undefined.
We can find the values of $x$ where $tan^2~x = 0$:
$tan^2~x = 0$
$tan~x = 0$
$\frac{sin~x}{cos~x} = 0$
$sin~x = 0$
$x = \pi~n,~~$ where $n$ is an integer
We can find the values of $x$ where $tan^2~x$ is undefined:
$cos~x = 0$
$x = \frac{\pi}{2}+\pi~n,~~$ where $n$ is an integer
Therefore, the function is discontinuous at the following points:
$x = \pi~n,~~$ where $n$ is an integer
$x = \frac{\pi}{2}+\pi~n,~~$ where $n$ is an integer
