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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: $$\tan\theta = \dfrac 3{\sqrt {91}} = \dfrac{3\sqrt{91}}{91}$$
