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?

• $\tan \theta = 0.4$ is positive in what Quadrants ? – AgentS Sep 1 '14 at 8:19
• The first and the third quadrants. – user170171 Sep 1 '14 at 8:20
• 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? – user170171 Sep 1 '14 at 8:21
• Exactly! Exactly! – AgentS Sep 1 '14 at 8:22
• 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. – Deuteu Sep 1 '14 at 8:24

$\tan\theta=0.4$

$\implies\tan^2\theta=0.16$

$\implies1+\tan^2\theta=1.16$

$\implies\sec^2\theta=1.16$

$\implies\sec\theta=\sqrt{1.16}$

$\implies\cos\theta=\pm\frac{1}{1.077}$

$\implies\cos\theta=\pm0.92$

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

Or, $\theta=156.92^\circ$

The cosine is negative in the second quadrant.

• 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. – user170171 Sep 1 '14 at 10:14
• 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. – 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).