Question

tan theta=12/5 and sec theta=-13/5

Answer 1

Tangent is producing a positive value, while secant is producing a negative value. This is telling us that we're in the `Third Quadrant`.

This will be important so we can label the sides of our triangle correctly.

Answer 2

\[\large \sec \theta=\frac{1}{\cos \theta}\]This is telling us that \(\large \cos\theta=-\dfrac{5}{13}\).

The reason this is important is because, the cosine represents a distance along the x-axis. See how it's negative? So we move to the left.

Recall this identity?\[\large \tan \theta=\frac{\sin \theta}{\cos \theta}\]How can the tangent be positive if we determined that the cosine is negative??
It would mean the sine has to be negative also right?\[\large \tan \theta=\frac{\sin \theta}{\cos \theta}=\frac{-\sin \theta}{-\cos \theta}\]This is what's telling us that we're in the 3rd quadrant. The fact that our x (cosine) and y (sine) have to be negative.

Understand that part? :)

Answer 3

yes.

Answer 4

So here is our triangle in the 3rd quadrant, with our angle \(\theta\)- Which I'm calling \(\theta'\) just to be a little more accurate. But don't worry about that. :)

Using our Trig Identities that relate sides, we can setup this triangle.

Answer 5

We determined earlier that both the cosine and sine need to be negative.
And tangent is \(\dfrac{opposite}{adjacent}\), so I labeled those sides accordingly.