# Finding indicated trigonometric value in specified quadrant

If I have csc $\theta$ = - $\dfrac{10}{3}$ and have to find tan $\theta$ in quadrant III, would I use 1 + $\cot^2\theta$ = $\csc^2\theta$ then find reciprocal which would be tan $\theta$?

If so, I get $\dfrac{3\sqrt{91}}{91}$ as tan, but that doesn't seem right as if I wanted to get cot $\theta$ from tan now, it would be different. I hope that made sense. Any help would be appreciated!

$$\csc \theta = \frac 1{\sin\theta} = -\frac {10}{3} \implies \sin\theta = -\frac 3{10} = \frac{\text{opposite}}{\text{hypotenuse}}$$

$$\tan \theta = \dfrac{\sin\theta}{\cos\theta} = \frac{\text{opposite}}{\text{adjacent}}$$

All you need is to find $\cos\theta$ in the third quadrant to compute tangent. Use the right angle that $\theta$ forms with the x-axis and the Pythagorean Theorem: $$3^2 + \text{adjacent}^2 = 10^2 \implies \text{adjacent} = \sqrt{91}$$ or else use the identity:

$$\sin^2\theta + \cos^2\theta = 1$$

knowing that $\cos \theta < 0$ in the third quadrant.

However, you are also correct having used your method: $$\tan\theta = \dfrac 3{\sqrt {91}} = \dfrac{3\sqrt{91}}{91}$$

• Oh wow, I was actually correct. I tried using the theorem you gave me, but I kept getting a positive answer. I get sqrt(91)/10. When I move 9/100 over to other side I subtracted it by 100/100 then took square root. Am I doing basic math wrong? D:
– o.o
Dec 6, 2013 at 13:30
• $\cos \theta$, in Quadrant III, is negative, (remember, the identity above is in $\cos^2 \theta$, so $\cos\theta = \pm \sqrt {\cos^2\theta}$. So $\cos\theta = -\dfrac{\sqrt{91}}{10}$, giving us $$\tan\theta = \dfrac{\sin\theta}{\cos\theta} = \dfrac{-3/10}{-\sqrt{91}/10} = \dfrac 3{\sqrt{91}} \cdots$$ Dec 6, 2013 at 13:34
• Oh I see. Thank you so much for your help! Makes a lot more sense now.
– o.o
Dec 6, 2013 at 13:36
• And no, you're not doing basic math wrong, it's just a matter of knowing which quadrant you're working in, and the corresponding signs of the trig functions. You're welcome! Dec 6, 2013 at 13:36
• @amWhy: Needs a TU w/G! +1 Dec 7, 2013 at 0:14