It is important to note first the following:
[tex]\tan \theta=\frac{y}{x}[/tex]and
[tex]\sec \theta=\frac{r}{x}[/tex]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.
[tex]\begin{gathered} \tan \theta=\frac{y}{x}>0 \\ \frac{+y}{+x}>0 \\ \frac{-y}{-x}>0 \end{gathered}[/tex]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.
[tex]\begin{gathered} \sec \theta=\frac{r}{x} \\ \frac{r}{x}>0 \\ \frac{r}{+x}>0 \end{gathered}[/tex]Between Quadrant I and Quadrant III, Quadrant I has a positive x value. Hence, θ is found in Quadrant I.
