Trigonometry 7th Edition

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

Chapter 1 - Section 1.5 - More on Identities - 1.5 Problem Set - Page 45: 34

Answer

(a). $\frac{a^{2}-b^{2}}{ab}$ (b). $\frac{\sin^{2}\theta-\cos^{2}\theta}{\sin\theta\cos\theta}$

Work Step by Step

(a). Given expression is- $\frac{a}{b} - \frac{b}{a}$ = $\frac{a}{a}\times\frac{a}{b} - \frac{b}{a}\times\frac{b}{b}$ = $\frac{a^{2}}{ab} - \frac{b^{2}}{ab}$ = $\frac{a^{2}-b^{2}}{ab}$ (b). Given expression is- $\frac{\sin\theta}{\cos\theta} - \frac{\cos\theta}{\sin\theta}$ = $\frac{\sin\theta}{\sin\theta}\times\frac{\sin\theta}{\cos\theta} - \frac{\cos\theta}{\sin\theta}\times\frac{\cos\theta}{\cos\theta}$ = $\frac{\sin^{2}\theta}{\sin\theta\cos\theta} - \frac{\cos^{2}\theta}{\sin\theta\cos\theta}$ = $\frac{\sin^{2}\theta-\cos^{2}\theta}{\sin\theta\cos\theta}$
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.