Chapter 5.3, Problem 42E

To determine

To find: The values of the remaining five trigonometric functions for $\sec \theta =\sqrt{3};\text{ Quadrant IV}$

Answer to Problem 42E

  $\begin{array}{l}\sin \theta =\frac{\sqrt{6}}{3},\csc \theta =\frac{\sqrt{6}}{2}\\ \cos \theta =\frac{\sqrt{3}}{3},\sec \theta =\sqrt{3}\\ \tan \theta =\sqrt{2},\cot \theta =\frac{\sqrt{2}}{2}\end{array}$

Explanation of Solution

Given information: Suppose $\theta$ is an angle in standard position whose terminal sides lies in the Quadrant IV.

Calculation:

First, find the missing side.

  $\begin{array}{c}S=\sqrt{{\left(\sqrt{3}\right)}^2-{\left(1\right)}^2}\\ =\sqrt{3-1}\\ =\sqrt{2}\end{array}$

The missing side is $\sqrt{2}$

  $\begin{array}{l}\sin \theta =\frac{\sqrt{2}}{\sqrt{3}}=\frac{\sqrt{6}}{3},\csc \theta =\frac{1}{\sin \theta }=\frac{1}{\frac{\sqrt{6}}{3}}=\frac{\sqrt{6}}{2}\\ \cos \theta =\frac{1}{\sqrt{3}}=\frac{\sqrt{3}}{3},\sec \theta =\frac{1}{\cos \theta }=\sqrt{3}\\ \tan \theta =\frac{\sqrt{2}}{1}=\sqrt{2},\cot \theta =\frac{1}{\tan \theta }=\frac{1}{\sqrt{2}}=\frac{\sqrt{2}}{2}\end{array}$