## Trigonometry (11th Edition) Clone

Published by Pearson

# Chapter 2 - Review Exercises - Page 94: 23

#### Answer

$\displaystyle \frac{7}{2}$

#### Work Step by Step

reference angles: $\left[\begin{array}{lllll} Quadr.: & I & II & III & IV\\ \theta' & \theta & 180^{o}-\theta & \theta-180^{o} & 360^{o}-\theta \end{array}\right]$ $300^{o}$ is in quadrant IV, where secant is positive, reference angle: $360^{o}-300^{o}=60^{o}$ Function Values of Special Angles$: \sec 300^{o}=\sec 60^{o}=2$ $150^{o}$ is in quadrant II, where cosine is negative, reference angle: $180^{o}-150^{o}=30^{o}$ Function Values of Special Angles: $\displaystyle \cos 240^{o}=-\cos 30^{o}=-\frac{\sqrt{3}}{2}$ $\tan 4$is in quadrant I, $\tan 45^{o}=1$ $\sec^{2}300^{o}-2\cos^{2}150^{o}+\tan 45^{o}=$ $=2^{2}-2(-\displaystyle \frac{\sqrt{3}}{2})^{2}+1$ $=4-2(\displaystyle \frac{3}{4})+1$ $=4-\displaystyle \frac{3}{2}+1$ $=\displaystyle \frac{7}{2}$

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