Algebra and Trigonometry 10th Edition

$\ln|tan~x|-\ln(1-cos^2x)=\ln|csc~x~sec~x|$
$\ln|tan~x|-\ln(1-cos^2x)=\ln\frac{|tan~x|}{1-cos^2x}=\ln\frac{|\frac{sin~x}{cos~x}|}{sin^2x}=\ln|\frac{1}{sin~x~cos~x}|=\ln|csc~x~sec~x|$ Notice that $sin^2x=|sin^2x|$