Trigonometry (11th Edition) Clone

Published by Pearson
ISBN 10: 978-0-13-421743-7
ISBN 13: 978-0-13421-743-7

Chapter 5 - Trigonometric Identities - Section 5.1 Fundamental Identities - 5.1 Exercises - Page 201: 46



Work Step by Step

$$\sec\theta=\frac{x+4}{x}$$ According to the Pythagorean Identity $$\tan^2\theta+1=\sec^2\theta$$ we can rewrite as $$\tan^2\theta=\sec^2\theta-1$$ Then, $$\tan\theta=\sqrt{\tan^2\theta}$$ This is the way we would be able to find an expression of $x$ for $\tan\theta$ 1) Find $\sec^2\theta$ $$\sec^2\theta=\frac{(x+4)^2}{x^2}$$ 2) Find $\tan^2\theta$ $$\tan^2\theta=\sec^2\theta-1$$ $$\tan^2\theta=\frac{(x+4)^2}{x^2}-1$$ $$\tan^2\theta=\frac{(x+4)^2-x^2}{x^2}$$ $$\tan^2\theta=\frac{(x+4-x)(x+4+x)}{x^2}$$ (for $a^2-b^2=(a-b)(a+b)$) $$\tan^2\theta=\frac{4(2x+4)}{x^2}$$ $$\tan^2\theta=\frac{8(x+2)}{x^2}$$ 3) Find $\tan\theta$ $$\tan\theta=\sqrt{\tan^2\theta}$$ $$\tan\theta=\frac{\sqrt{8(x+2)}}{\sqrt{x^2}}$$ $$\tan\theta=\frac{2\sqrt{2(x+2)}}{|x|}$$ (since $\sqrt{a^2}=|a|$) Unfortunately, we cannot eliminate the absolute value $||$ sign, since there is not enough information to know whether $x\gt0$ or $\lt0$.
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.