Since tan^2(x)>0 for all x, there is no need to worry about any negative under the natural logarithm function. When tan^2(x) = 0 and tan^2(x) is undefined
Since tan^2(x) = sin^2(x)/cos^2(x), we see that any problem will result when either the numerator (tan^2(x)=0) or denominator equals zero (sin^2(x)=0 or cos^2(x)=0)
Solve the two equations gives:
sin(x)=0 ==> x=Ď€k for integer k
cos(x)=0 ==> x=Ď€/2 +/- 2Ď€k and x=3Ď€/2 +/- 2Ď€k integer k
Combing all three gives discontinuities at x=+/- πk/2 for all integers k
