Chapter 1.6, Problem 10E

To determine

To find: Thevalue of other trigonometric functions if $\tan \theta =-1$ such that $\text{sin}\theta <0$ .

Answer to Problem 10E

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

Explanation of Solution

Given information:The given condition is $\tan \theta =-1,\text{ sin}\theta <0$ .

Calculation:

The value of secant function can be calculated as:

  $\begin{array}{c}\sec \theta =\sqrt{1+{\tan }^2\theta }\\ =\sqrt{1+{\left(-1\right)}^2}\\ =\sqrt{2}\end{array}$

The value of cosine function can be calculated as:

  $\begin{array}{c}\cos \theta =\frac{1}{\sec \theta }\\ =\frac{1}{\sqrt{2}}\\ =\frac{\sqrt{2}}{2}\end{array}$

The value of sine function can be calculated as:

  $\begin{array}{c}\sin \theta =\sqrt{1-{\cos }^2\theta }\\ =\sqrt{1-{\left(\frac{1}{\sqrt{2}}\right)}^2}\\ =-\frac{1}{\sqrt{2}}\\ =-\frac{\sqrt{2}}{2}\end{array}$

The value of cosecant function can be calculated as:

  $\begin{array}{c}\text{csc}\theta =\frac{1}{\sin \theta }\\ =\frac{1}{-\frac{\sqrt{2}}{2}}\\ =-\frac{2}{\sqrt{2}}\\ =-\sqrt{2}\end{array}$

The value of cotangent function can be calculated as:

  $\begin{array}{c}\text{cot}\theta =\frac{1}{\tan \theta }\\ =\frac{1}{-1}\\ =-1\end{array}$

Therefore, the other trigonometric functions are $\sin \theta =-\frac{\sqrt{2}}{2}$ , $\cos \theta =\frac{\sqrt{2}}{2}$ , $\text{csc}\theta \text{=}-\sqrt{2}$ , $\sec \theta =\sqrt{2}$ and $\cot \theta =-1$ .