Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter 7 - Trigonometric Identities and Equations - Quiz - Page 683: 7


$\dfrac {1+\sin \theta }{\cot ^{2}\theta }=\dfrac {\sin \theta }{csc\theta -1}$

Work Step by Step

$\dfrac {1+\sin \theta }{\cot ^{2}\theta }=\dfrac {\left( 1+\sin \theta \right) }{\dfrac {\cos ^{2}\theta }{\sin ^{2}\theta }}=\dfrac {\sin ^{2}\theta \left( 1+\sin \theta \right) }{1-\sin ^{2}\theta }\dfrac {=\sin ^{2}\theta \left( 1+\sin \theta \right) }{\left( 1-\sin \theta \right) \left( 1+\sin \theta \right) }=\dfrac {\sin ^{2}\theta }{1-\sin \theta }=\dfrac {\sin \theta }{\dfrac {1}{\sin \theta }-1}=\dfrac {\sin \theta }{csc\theta -1}$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.