Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 1 - Functions - 1.4 Trigonometric Functions and Their Inverse - 1.4 Exercises: 30

Answer

\[\tan \theta = \frac{{\sin \,\theta }}{{\cos \theta }}\]

Work Step by Step

\[\begin{gathered} Definition\,:\,trigonometric\,functions \hfill \\ \hfill \\ Let\,P\,\left( {x,y} \right)\,be\,a\,point\,on\,a\,circle\,\,\,with\,radius\,r\,\,associated \hfill \\ with\,the\,angle\,\,\theta .\, \hfill \\ \hfill \\ therefore \hfill \\ \hfill \\ \sin \,\theta = \frac{y}{r}\,,\,\cos \theta = \frac{x}{r}\,\,,\,\tan \theta = \frac{y}{x} \hfill \\ \hfill \\ \cot \theta = \frac{x}{y}\,\,,\,\sec \theta = \frac{r}{x}\,\,,\,\csc \,\theta = \frac{r}{y} \hfill \\ \hfill \\ Using\,\,definition \hfill \\ \hfill \\ \tan \theta = \frac{y}{x} = \frac{y}{x} \cdot \frac{{\frac{{\frac{1}{r}}}{1}}}{r} = \frac{{\frac{{\frac{y}{r}}}{x}}}{r} \hfill \\ \hfill \\ or \hfill \\ \hfill \\ \tan \theta = \frac{{\sin \,\theta }}{{\cos \theta }} \hfill \\ \end{gathered} \]
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