# What is the domain of y=-tan^2(x) ?

Jul 23, 2018

$x \ne \left(2 k + 1\right) \frac{\pi}{4} , k = 0 , \pm 1 , \pm 2 , \pm 3. \ldots$. See graph.

#### Explanation:

The asymptotes are x = zeros of the denominator $\cos 2 x$,

$x = \left(2 k + 1\right) \frac{\pi}{4} , k = 0 , \pm 1 , \pm 2 , \pm 3. \ldots$
So, $x \ne \left(2 k + 1\right) \frac{\pi}{4} , k = 0 , \pm 1 , \pm 2 , \pm 3. \ldots$

graph{(x-pi/4+0.01y)(y-tan(2x)) =0[-4 4 -2 2]}

One asymptote $x = \frac{\pi}{4}$ is marked.

The space in between the asympyote and the nearby branch of the

graph $\downarrow$ as $\left\mid y \right\mid \uparrow$ but never becomes 0. So this x is not

in the graph..