Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 4 - Section 4.3 - Right Triangle Trigonometry - Exercise Set - Page 561: 51



Work Step by Step

Since, $\tan( \dfrac{\pi}{2}-\theta)=\cot \theta $ The trigonometric ratios are as follows: $\sin \theta =\dfrac{y}{r} \\ \cos \theta =\dfrac{x}{r} \\ \tan \theta =\dfrac{y}{x}\\ \csc \theta =\dfrac{r}{y} \\ \sec \theta =\dfrac{r}{x} \\ \cot \theta =\dfrac{x}{y}$ where, $ r=\sqrt {x^2+y^2}$ Now, we have $\tan( \dfrac{\pi}{2}-\theta)=\cot \theta=\dfrac{1}{4}$
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