## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter 4 - Section 4.2 - Trigonometric Functions: The Unit Circle - Exercise Set - Page 548: 28

#### Answer

tan t = $\frac{2\sqrt 5}{5}$ csc t =$\frac{3}{2}$ sec t =$\frac{3\sqrt 5}{5}$ cot t = $\frac{\sqrt 5}{2}$

#### Work Step by Step

tan t is computed by $\frac{sin t}{cos t}$ tan t = $\frac{sin t}{cos t}$ =$\frac{\frac{2}{3}}{\frac{\sqrt 5}{3}}$ =$\frac{2}{\sqrt 5}$*$\frac{\sqrt 5}{\sqrt 5}$ = $\frac{2\sqrt 5}{5}$ csc t = $\frac{1}{sin t}$ = $\frac{1}{\frac{2}{3}}$ =$\frac{3}{2}$ sec t = $\frac{1}{cos t}$ = $\frac{1}{\frac{\sqrt 5}{3}}$ = $\frac{3}{\sqrt 5}$ * $\frac{\sqrt 5}{\sqrt 5}$ = $\frac{3\sqrt 5}{5}$ cot t = $\frac{1}{tan t}$ = $\frac{1}{\frac{2}{\sqrt 5}}$ = $\frac{\sqrt 5}{2}$

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