# Thread: Triganomic Ratios

1. ## Triganomic Ratios

Find sec(theta) if tan(theta)=sqrt(3)/3, and sin(theta)<0.

2. Originally Posted by Chinnie15
Find sec(theta) if tan(theta)=sqrt(3)/3, and sin(theta)<0.
use the identity ...

$sec^2{t} = 1 + \tan^2{t}$

3. Thanks!

4. Originally Posted by Chinnie15
Thanks!

I'm not sure how to find t? I've been trying all day to learn this stuff, and it's just not clicking.
$\sec^2{t} = 1 + \left(\frac{\sqrt{3}}{3}\right)^2$

solve for $\sec{t}$ ... mind your signs.

5. Ok, I understand the rest now, but what is the sin<theta? What does that have to do with the problem?

I took the root of both sides and have ended up with sqrt(3)+1/3. Is this correct?

6. Originally Posted by Chinnie15
Ok, I understand the rest now, but what is the sin<theta? What does that have to do with the problem?
you were given $\tan{t} = \frac{\sqrt{3}}{3}$ ... a (+) value

you were also told $\sin{\theta} < 0$ ... a (-) value

those two pieces of information tell you what quadrant $\theta$ is in ... so you can determine the correct sign for $\sec{\theta}$

7. Ok, thanks!

I have 2/sq(3)=secΘ

8. Find sec(theta) if tan(theta)=sqrt(3)/3, and sin(theta)<0.

tan is + and sin is < 0, that is - ; that is in the 3rd quadrant.

But in the 3rd quadrant only tan and cot are positive, the rest negative.

, from this we have,
sec = -(2/3)(sqrt 3)

the negative sign ( - ) is introduced since it is in the 3rd quadtant