<s>[INST] Given $\csc\theta=-\frac53$ and $\pi<\theta<\frac32\pi$, evaluate sine ,cosine, and tangent of $2\theta$ If $\csc\theta=\frac{-5}{3}$, what is the exact value of $\tan(2\theta)$, $\sin(2\theta)$, and $\cos(2\theta)$ on the interval of $\left(\pi, \frac{3\pi}{2}\right)$?
I think I'm getting the fraction negatives wrong. I've used the sine formula($2\sin{\theta}\cos{\theta}$), the cosine formula ($2\cos^2{\theta}-1$) and the tangent formula, $\left(\frac{\sin(2\theta)}{\cos(2\theta)}\right)$
originally answering the problem I got $\sin(2\theta)=\frac{24}{25}$,
 $\cos(2\theta)=\frac{7}{25}$, and $\tan(2\theta)=\frac{24}{7}$
 [/INST] Notice that $$\csc{x} = \frac{1}{\sin{x}}$$
and so $$\csc {x} = \frac{-5}{3} = \frac{1}{\sin{x}}$$ and so $\sin{x} = \frac{-3}{5}$
now since $\color{blue}{\sin^2(x) + \cos^2(x) = 1}$ then we have that $$\big(\frac{-3}{5}\big)^2 + \cos^2(x) =  1$$ and so we get that $$\frac{9}{25} + \cos^2(x) = 1$$ and so $\cos^2{x} = 1 - \frac{9}{25} = \frac{16}{25}$ and so $$\cos{x} = \pm \frac{4}{5}$$
Now you want to find $\sin{2x}$ then you should use the identity $\color{red}{\sin{2x} = 2\sin{x}\cos{x}}$ and so you get $$\sin{2x} = 2\times {\frac{-3}{5}} \times \pm \frac{4}{5}$$.
Now you should the use other identity $\color{purple}{\cos{2x} = \cos^2{x} - \sin^2{x}}$ to find $\cos{2x}$ and finally you should use the identity that $$\color{green}{\tan{2x} = \frac{\sin{2x}}{\cos{2x}}}$$
To finish up your question.
I advise you to take a look at that link to study your identities 
Double angle indentities
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