# Find the exact value of $\sec \theta$ given the value of $\tan \theta$

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.

• It helps to remember that $\sec^2\theta=1+\tan^2\theta$. Of course, you need to be careful about the sign. So remember the quadrant rule. Commented Sep 24, 2019 at 18:02

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 $$\sec(\theta)=-\sqrt{50}=-5\sqrt{2}$$.

Since $$\tan\theta >0$$, the angle $$\theta$$ must be in the third quadrant, over which secant function takes negative values.

Thus,

$$\sec\theta =- \sqrt{1+\tan^2\theta}=- \sqrt{1+49}= -5\sqrt 2$$

The idea here is that if $$\tan\theta=7$$, $$\theta\in(-\pi,0)$$, this tells us that the angle is in the third quadrant, and as such, $$\sec\theta$$ has to be negative. You are on the right track, in general, but the correct answer would be $$\boxed{\sec\theta=-5\sqrt2}$$.