# Confused about which quadrant the solution to this trigonometry question is in

Below is the textbook solution to a question I'm confused about (areas of confusion in yellow).

I got $\tan\theta = \pm \frac{7}{24}$ whereas the correct answer is $\tan\theta = -\frac{7}{24}$.

I understand why $\theta$ is in the 4th quadrant if $\cos\theta$ is positive and $\theta$ is reflex. However, since $\tan\theta = \frac{7}{24}$ (and $\theta$ is reflex) doesn't this imply that $\theta$ is in both the 3rd and 4th quadrants, and therefore that $\tan\theta$ can be both positive and negative?

Thanks!

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You write "since $\tan\theta=\frac{7}{24}$", but that isn't anywhere in the solution. It says $\tan\phi=\frac{7}{24}$, but $\phi$ is not $\theta$. (I personally find this detour through a different angle more confusing than illuminating.) You may have been thrown off by $\cos\theta=\frac{24}{25}$, which might be a misprint, since there's no $\theta$ in the triangle (and there couldn't be, since $\theta$ is reflex); I suspect this was meant to say $\cos\phi=\frac{24}{25}$. – joriki Apr 15 '11 at 11:42
Using the identity $1 + \tan^2\theta = \sec^2\theta$ I deduced that $\tan\theta = \pm \frac{7}{24}$ after some manipulation. I guess my question is, why can't I use this to determine which quadrant(s) $\theta$ is in, since I deduced the fact from the original $\cos\theta = \frac{24}{25}$? – Danny King Apr 15 '11 at 12:00
Because you can't determine the sign of $\tan\theta$ from an equation in which only its square occurs; the square doesn't contain any information about the sign. You seem to be slightly misusing or misunderstanding the notation $\tan\theta=\pm\frac{7}{24}$. It doesn't mean that $\theta$ can take both of those values; it only summarizes what you know from that one equation (namely that $\theta$ can only take one of those two values) but doesn't preclude the possibility that you can determine the sign from other information; in this case, from the information that the cosine is positive. – joriki Apr 15 '11 at 12:22
As joriki says, given that the cosine is positive, the angle must be in the 1st quadrant or the 4th; given that the angle is reflex, we can forget about the 1st quadrant. So, it's in the 4th, whence the tangent is negative. – Gerry Myerson Apr 15 '11 at 12:47
Thank you both, that has cleared it up. Would either of you like to post this an an answer so that I can accept it? If not I shall post my own answer. – Danny King Apr 15 '11 at 13:48

You write "since $\tan\theta=\frac{7}{24}$", but that isn't anywhere in the solution. It says $\tan\phi=\frac{7}{24}$, but $\phi$ is not $\theta$. (I personally find this detour through a different angle more confusing than illuminating.) You may have been thrown off by $\cos\theta=\frac{24}{25}$, which might be a misprint, since there's no $\theta$ in the triangle (and there couldn't be, since $\theta$ is reflex); I suspect this was meant to say $\cos\phi=\frac{24}{25}$.

Answer to the further question why $1+\tan^2\theta=\sec^2\theta$ can't be used to determine which quadrant(s) $\theta$ is in:

Because you can't determine the sign of tanθ from an equation in which only its square occurs; the square doesn't contain any information about the sign. You seem to be slightly misusing or misunderstanding the notation $\tan\theta=\pm\frac{7}{24}$. It doesn't mean that $\theta$ can take both of those values; it only summarizes what you know from that one equation (namely that $\theta$ can only take one of those two values) but doesn't preclude the possibility that you can determine the sign from other information; in this case, from the information that the cosine is positive.

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If $0^{\circ} \lt \varphi \le 180^{\circ}$, then the reflex angle of $\varphi$ is $\theta = 360^{\circ} - \varphi.$

If $\cos \theta = \dfrac{24}{25}$ and $\theta$ is a reflex angle, then $\theta$ is in quadrant IV.

So $\sin \theta$ is $-\dfrac{7}{25}$.

So $\tan \theta = \dfrac{\sin \theta}{\cos \theta} = \dfrac{-7}{24}$

It looks like you found a typo.

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