If $\tan \theta = 7$ and $-\pi < \theta < 0$ then what is the exact value of $\sec \theta?$
I tried doing $\tan \theta = \frac{y}{x}= 7 $, so $y = 7x$ and subbed that into $1 = \sqrt{y^2 + x^2}$, however I ended up getting $x = \frac{1}{5\sqrt{2}}$ and thus $\sec \theta = -5\sqrt{2}$ however this is incorrect.
EDIT: sorry I accidentally left out the negative sign and I did indeed consider the quadrants, I actually got the right answer but I stupidly thought I didn't since the correct answer was listed as $-\frac{10}{\sqrt{2}}$. For some reason it did not occur to me to rationalize it. However thanks for your responses since when I saw that you guys had the same answer as me it made me go back and look at the "correct" answer listed, so now I realized I simply overlooked something really silly.
Best Answer
You can use the identity $1+\tan^2\theta=\sec^2{\theta}$.
Since you know that $\displaystyle -\pi < \theta < -\frac{\pi}{2}$ from the sign of $\tan$, you can determine that $\sec$ is negative.
Now, you have that $\sec(\theta)=-\sqrt{50}=-5\sqrt{2}$.