Chapter 4.4, Problem 46E

To determine

The trigonometric function $\text{tan}\left(-\frac{\pi }{2}\right)$ of the quadrant angle.

Answer to Problem 46E

The trigonometric function $\text{tan}\left(-\frac{\pi }{2}\right)$ of the quadrant angle is $\text{not defined}$

Explanation of Solution

Given info:

  $\theta =-\frac{\pi }{2}$

Formula used:

  

  $\begin{array}{l}(x,y)\text{ for angle 0 = (1,0)}\\ (x,y)\text{ for angle 90 = (0,1)}\\ (x,y)\text{ for angle 180 = (-1,0)}\\ (x,y)\text{ for angle 270 = (0,-1)}\end{array}$

  $\begin{array}{l}r=\sqrt{x^2+y^2}\\ \text{sin}\theta \text{=}\frac{y}{r}\\ \text{cos}\theta \text{=}\frac{x}{r}\\ \text{tan}\theta \text{=}\frac{y}{x}\\ \text{csc}\theta \text{=}\frac{r}{y}\\ \text{sec}\theta \text{=}\frac{r}{x}\\ \text{cot}\theta \text{=}\frac{x}{y}\end{array}$

Calculation:

Here we need to find $\text{tan}\left(-\frac{\pi }{2}\right)$

So,

We know

  $\tan \left(-\theta \right)=-\tan \theta$

  $\begin{array}{l}(x,y)\text{ for angle }\frac{\pi }{2}\text{ = (0,1)}\\ x=0\\ y=1\\ r=\sqrt{x^2+y^2}\\ r=\sqrt{0^2+1^2}=1\\ \text{tan}\theta \text{=}\frac{y}{x}\\ \text{tan}\left(-\frac{\pi }{2}\right)\text{=-tan}\frac{\pi }{2}=-\frac{1}{0}\text{= not defined}\end{array}$

Conclusion:

Thus,thetrigonometric function $\text{tan}\left(-\frac{\pi }{2}\right)$ of the quadrant angle is $\text{not defined}$