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

Question

Madelyn Adams · Accepted Answer

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

Adam Aguirre · Answer

undefined To find the domain and range of $ y = \ln(\tan^2(x)) $:

1. **Domain:**
   - $ \tan(x) $ is defined for all real numbers except odd multiples of $ \frac{\pi}{2} $ (i.e., $ x $ cannot be $ \frac{(2n+1)\pi}{2} $ where $ n $ is an integer).
   - $ \tan^2(x) $ is always non-negative since it squares the tangent function.
   - $ \ln(u) $ is defined only for positive $ u $.
   - Therefore, the domain of $ y = \ln(\tan^2(x)) $ is $ x $ such that $ x $ is not an odd multiple of $ \frac{\pi}{2} $.

2. **Range:**
   - The range of $ y = \ln(\tan^2(x)) $ is all real numbers since $ \tan^2(x) $ can take any non-negative value, and $ \ln(u) $ is defined for all positive $ u $.