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?

  • 1
    $\begingroup$ $\tan \theta = 0.4$ is positive in what Quadrants ? $\endgroup$ – AgentS Sep 1 '14 at 8:19
  • $\begingroup$ The first and the third quadrants. $\endgroup$ – user170171 Sep 1 '14 at 8:20
  • 1
    $\begingroup$ Right, so that means that to satisfy tanθ being positive, then the answer is limited to the third quadrant only because the 1st has been dismissed. Is that correct? $\endgroup$ – user170171 Sep 1 '14 at 8:21
  • $\begingroup$ Exactly! Exactly! $\endgroup$ – AgentS Sep 1 '14 at 8:22
  • $\begingroup$ I think you forgot to get rid of the square in the last line. Just a mistake with no consequences on your question but just to clarify it. $\endgroup$ – Deuteu Sep 1 '14 at 8:24








So, either $\theta=23.07^\circ$

Or, $\theta=156.92^\circ$

The cosine is negative in the second quadrant.

  • $\begingroup$ The question did not ask for the value of theta and your answer does not explain why the third and fourth quadrants are not valid. $\endgroup$ – user170171 Sep 1 '14 at 10:14
  • $\begingroup$ Until and unless you calculate $\theta$ you can't guarantee in which quadrant it will fall. If you measure $156.92^\circ$ it falls in the second quadrant. $\endgroup$ – MonK Sep 1 '14 at 10:49

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).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy