Trigonometry 7th Edition

Published by Cengage Learning
ISBN 10: 1111826854
ISBN 13: 978-1-11182-685-7

Chapter 1 - Section 1.4 - Introduction to Identities - 1.4 Problem Set - Page 40: 46


$\cos\theta$ = $\frac{5}{13}$ $\tan\theta$ = $\frac{12}{5}$ $\csc\theta$ = $\frac{13}{12}$ $\sec\theta$ = $\frac{13}{5}$ $\cot\theta$ = $\frac{5}{12}$

Work Step by Step

We know from first Pythagorean identity that- $\cos\theta$ = ± $\sqrt (1-\sin^{2}\theta)$ As $\theta$ terminates in Q I, Therefore $\cos\theta$ will be positive- $\cos\theta$ = $\sqrt (1-\sin^{2}\theta)$ substitute the given value of $\cos\theta$- $\cos\theta$ = $\sqrt (1-(\frac{12}{13})^{2})$ $\cos\theta$ = $\sqrt (1-\frac{144}{169})$ $\cos\theta$ = $\sqrt (\frac{169-144}{169})$ = $\sqrt (\frac{25}{169})$ $\cos\theta$ = $\frac{5}{13}$ By ratio identity- $\tan\theta$ = $\frac{\sin\theta}{\cos\theta}$ Substituting the values of $\sin\theta$ and $\cos\theta$- $\tan\theta$ = $\frac{12/13}{5/13}$ = $\frac{12}{5}$ From reciprocal identities- $\csc\theta$ = $\frac{1}{\sin\theta}$ = $\frac{1}{12/13}$ = $\frac{13}{12}$ $\sec\theta$ = $\frac{1}{\cos\theta}$ = $\frac{1}{5/13}$ = $\frac{13}{5}$ $\cot\theta$ = $\frac{1}{\tan\theta}$ = $\frac{1}{12/5}$ = $\frac{5}{12}$ As $\theta$ terminates in Q I, Therefore all trigonometric functions will be positive.
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