Question

Let theta be an angle in standard position. Name the quadrant in which theta lies. tan (theta) > 0, sec (theta) > 0options:IIIIIIIVPlease help me find the answer and explain how to find it in the future

Answer 1

It is important to note first the following:

`\tan \theta=(y)/(x)`

and

`\sec \theta=(r)/(x)`

The first condition given in the question is tan θ > 0 hence, we can assume that both x and y are either positive or negative so that we can have a positive number.

`\begin{gathered} \tan \theta=(y)/(x)>0 \\ (+y)/(+x)>0 \\ (-y)/(-x)>0 \end{gathered}`

This means our θ falls either in Quadrant I or Quadrant III because the coordinate in Quadrant I are (+, +) while in Quadrant III is (-, -).

Now, the second condition given is that sec θ > 0, hence, we can assume that the value of "x" has to be positive so that sec θ will be greater than zero.

`\begin{gathered} \sec \theta=(r)/(x) \\ (r)/(x)>0 \\ (r)/(+x)>0 \end{gathered}`

Between Quadrant I and Quadrant III, Quadrant I has a positive x value. Hence, θ is found in Quadrant I.