Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Appendix D - Trigonometry - D Exercises: 49

Answer

$cot^{2}\theta+sec^{2}\theta=tan^{2}\theta+csc^{2}\theta$

Work Step by Step

Need to prove the identity $cot^{2}\theta+sec^{2}\theta=tan^{2}\theta+csc^{2}\theta$ In order to prove this, take left side isolate. $cot^{2}\theta+sec^{2}\theta=(csc^{2}\theta-1)+(tan^{2}\theta-1)$ $cot^{2}\theta+sec^{2}\theta=csc^{2}\theta+tan^{2}\theta$ Hence, $cot^{2}\theta+sec^{2}\theta=tan^{2}\theta+csc^{2}\theta$
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