Precalculus (6th Edition)

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

Chapter 7 - Trigonometric Identities and Equations - Summary Exercises Verifying Trigonometric Identities - Exercises: 17

Answer

$\dfrac {\tan ^{2}t+1}{\tan tcsc^{2}t}=\tan t$

Work Step by Step

$\dfrac {\tan ^{2}t+1}{\tan tcsc^{2}t}=\dfrac {\dfrac {\sin ^{2}t}{\cos ^{2}t}+1}{\dfrac {\sin t}{\cos t}\dfrac {1}{\sin ^{2}t}}=\dfrac {\dfrac {1}{\cos ^{2}t}}{\dfrac {1}{\cos t\sin t}}=\dfrac {\cos t\sin t}{\cos ^{2}t}=\dfrac {\sin t}{\cos t}=\tan t$
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