Trigonometry - Answers in Multiple Quadrants May I please get some help with this question?
Find the value of $\sec \theta $ if $\tan \theta = 0.4$ and $\theta$ is not in the 1st quadrant.
Now, here is my current working:
$\frac{\pi}{2} < \theta < 2\pi$
$\tan^2 \theta + 1 = \sec^2 \theta$
$0.4^2 + 1 = \sec^2 \theta$
$\frac{29}{25} = \sec^2 \theta $
$\pm \frac{\sqrt{29}}{5} = \sec \theta $
Now here's where I'm stuck. My textbook tells me that the answer is $\frac{-\sqrt{29}}{5}$. But how can that be if $\cos\theta$ is positive in the the 4th quadrant (not 1st quadrant must mean that that it is 2nd, 3rd and 4th quadrant).
Why is the answer only a negative?
 A: $\tan\theta=0.4$
$\implies\tan^2\theta=0.16$
$\implies1+\tan^2\theta=1.16$
$\implies\sec^2\theta=1.16$
$\implies\sec\theta=\sqrt{1.16}$
$\implies\cos\theta=\pm\frac{1}{1.077}$
$\implies\cos\theta=\pm0.92$
So, either $\theta=23.07^\circ$
Or, $\theta=156.92^\circ$
The cosine is negative in the second quadrant.
A: Think of it this way, $\tan\theta\gt 0$ whenever the horizontal and negative distances of a point on the terminal side of an angle in standard position are either both positive or both negative. This means that the terminal side of the angle $\theta$ either lies in the 1st or 3rd quadrants if $\tan\theta>0$. It may lie in the 1st quadrant because both the horizontal and vertical distances from the origin (aka the x-values and y-values) of points on the terminal side are both positive; it may lie in the 4th quadrant because both distances (or x-value and y-values if you prefer) are negative. But the problem statement discounts the 1st quadrant, so $\theta$ must be in the 4th quadrant.
In the 4th quadrant, $\sec\theta=\frac{\text{absolute distance from the origin}}{\text{horizontal distance from the origin}}$. If we consider the absolute distance from the origin as a positive quantity, and (as has already been determined) the horizontal distance from the origin as a negative quantity, then $\sec\theta<0$, thus $\sec\theta = -\frac{\sqrt{29}}{5}$.
Disclaimer: I am aware that this simplistic description doesn't hold when the absolute distance from the origin (aka the radius) is negative, in which case a 3rd quadrant angle flips the terminal side back into the 1st quadrant, giving positive $x$ and $y$ values, but $\sec\theta$ still evaluates correctly (due to the negative radius).
