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!