Question

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

Answer 1

Asymptotic `x ne k pi darr and ne (2k + 1)pi/2 uarr, k = 0, +-1, +-2, +-3, ...`

Explanation:

`y = ln ((tan x )^2) = 2 ln tan x` is real when

`x ne k pi , ne (2k + 1)pi/2 and notin Q_2` or `Q_4`,

to negate respectively, `sin x = 0 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 oo, x to ( 2k + 1 )pi/2 and - oo, x to kpi`.

It is delightful yo observe alternate asymptoticity, by sliding the graph `uarr and darr`. As you just see, it is in hiding.