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

1 Answer
Jul 29, 2018

Answer:

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.