Calculus: Early Transcendentals 9th Edition

Published by Cengage Learning
ISBN 10: 1337613924
ISBN 13: 978-1-33761-392-7

APPENDIX D - Trigonometry - D Exercises - Page A 34: 55

Answer

$csc\phi$ $+cot\phi$ $=\frac{sin\phi}{1-cos \phi}$

Work Step by Step

We need to prove the identity $csc\phi$ $+cot\phi$ $=\frac{sin\phi}{1-cos \phi}$ Let us solve the left side of the given identity. $csc\phi$ $+cot\phi$ $=\frac{1}{sin\phi}+\frac{cos \phi}{sin\phi}$ $=\frac{1+cos \phi}{sin\phi}$ $=\frac{1+cos \phi}{sin\phi}\times \frac{1-cos \phi}{1-cos\phi}$ $=\frac{1-cos^{2} \phi}{sin\phi(1-cos\phi)}$ $=\frac{sin^{2} \phi}{sin\phi(1-cos\phi)}$ Hence, $csc\phi$ $+cot\phi$ $=\frac{sin\phi}{1-cos \phi}$
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