Answer
$\ln|tan~x|-\ln(1-cos^2x)=\ln|csc~x~sec~x|$
Work Step by Step
$\ln|tan~x|-\ln(1-cos^2x)=\ln\frac{|tan~x|}{1-cos^2x}=\ln\frac{|\frac{sin~x}{cos~x}|}{sin^2x}=\ln|\frac{1}{sin~x~cos~x}|=\ln|csc~x~sec~x|$
Notice that $sin^2x=|sin^2x|$
Algebra and Trigonometry 10th Edition
by Larson, Ron
Published by
Cengage Learning
ISBN 10:
9781337271172
ISBN 13:
978-1-33727-117-2
Chapter 7 - 7.1 - Using Fundamental Identities - 7.1 Exercises - Page 514: 63
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