# How do you find the domain and range of y=ln(tan^2(x))?

Jul 29, 2018

Asymptotic $x \ne k \pi \downarrow \mathmr{and} \ne \left(2 k + 1\right) \frac{\pi}{2} \uparrow , k = 0 , \pm 1 , \pm 2 , \pm 3 , \ldots$

#### Explanation:

$y = \ln \left({\left(\tan x\right)}^{2}\right) = 2 \ln \tan x$ is real when

$x \ne k \pi , \ne \left(2 k + 1\right) \frac{\pi}{2} \mathmr{and} \notin {Q}_{2}$ or ${Q}_{4}$,

to negate respectively, $\sin x = 0 \mathmr{and} \cos x = 0$.
Graph:
graph{(tan x - e^(y/2))(x)(x^2-1/4(pi)^2)(x^2-(pi)^2)(x^2-9/4(pi)^2)(x^2-4(pi)^2)(x^2-25/4(pi)^2)(x^2-9(pi)^2)=0}

Note that $y \to \infty , x \to \left(2 k + 1\right) \frac{\pi}{2} \mathmr{and} - \infty , x \to k \pi$.

It is delightful yo observe alternate asymptoticity, by sliding the graph $\uparrow \mathmr{and} \downarrow$. As you just see, it is in hiding.