Answer
$\dfrac {\sin ^{2}\theta }{1-\sin ^{2}\theta }$
Work Step by Step
$\tan ^{2}\theta =\dfrac {\sin ^{2}\theta }{\cos ^{2}\theta }=\dfrac {\sin ^{2}\theta }{1-\sin ^{2}\theta }$
Precalculus: Mathematics for Calculus, 7th Edition
by Stewart, James; Redlin, Lothar; Watson, Saleem
Published by
Brooks Cole
ISBN 10:
1305071751
ISBN 13:
978-1-30507-175-9
Chapter 6 - Review - Exercises - Page 529: 51
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