Chapter 12.9, Problem 7MRE

To determine

Find the five other trigonometric functions of $\left(\theta \right)$ for which $\tan \theta =-1\text{, }270°<\theta <360°$ .

Answer to Problem 7MRE

The five trigonometric function are written as,

  $\sin \theta =-\frac{\sqrt{2}}{2},\text{cos}\theta =\frac{\sqrt{2}}{2}\text{,cot}\theta =-1,\sec \theta =\sqrt{2}\text{ and csc}\theta =-\sqrt{2}$

Explanation of Solution

Given information: The given trigonometric function is,

  $\tan \theta =-1\text{, }270°<\theta <360°\mathrm{....}\left(1\right)$

Calculation: From (1) the angle $\left(\theta \right)$ lies in fourth quadrant therefore except cosine and secant functions other four will be negative in this quadrant. Therefore the values for the five trigonometric functions are evaluated as,

  $\begin{array}{c}\tan \theta =-1\text{, }270°<\theta <360°\mathrm{....}\left(\text{Given}\right)\\ \text{Therefore, }\cot \theta =-1\\ \sec \theta =\pm \sqrt{1+{\tan }^2\theta }\\ =\pm \sqrt{1+{\left(-1\right)}^2}\\ =\sqrt{2},\left\{270°<\theta <360°\right\}\\ \cos \theta =\frac{1}{\sec \theta }\\ =\frac{\sqrt{2}}{2}\end{array}$

Similarly,

  $\begin{array}{c}\csc \theta =\pm \sqrt{1+{\cot }^2\theta }\\ =\pm \sqrt{1+{\left(-1\right)}^2}\\ =-\sqrt{2},\left\{270°<\theta <360°\right\}\\ \sin \theta =\frac{1}{\csc \theta }\\ =\frac{1}{-\sqrt{2}}\\ =-\frac{\sqrt{2}}{2},270°<\theta <360°\end{array}$