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 $\tan(2\theta)=\frac{24}{7}$

We are given that $\csc\theta = -\dfrac{5}{3}$ with $\pi < \theta < \dfrac{3\pi}{2}$.  
Since $\csc\theta = \dfrac{1}{\sin\theta}$,
$$\sin\theta = \frac{1}{\csc\theta} = \frac{1}{-\frac{5}{3}} = -\frac{3}{5}$$
Using the trigonometric identity $\sin^2\theta + \cos^2\theta = 1$ yields
\begin{align*}
\cos^2\theta & = 1 - \sin^2\theta\\
             & = 1 - \left(-\frac{3}{5}\right)^2\\
             & = 1 - \frac{9}{25}\\
             & = \frac{16}{25}
\end{align*}
Since $\pi < \theta < \dfrac{3\pi}{2}$, $\theta$ is a third-quadrant angle, so we take the negative square root.  Thus,
$$\cos\theta = -\frac{4}{5}$$
Hence, 
\begin{align*}
\sin(2\theta) & = 2\sin\theta\cos\theta\\
              & = 2\left(-\frac{3}{5}\right)\left(-\frac{4}{5}\right)\\
              & = \frac{24}{25}\\
\cos(2\theta) & = \cos^2\theta - \sin^2\theta\\
              & = \left(-\frac{4}{5}\right)^2 - \left(-\frac{3}{5}\right)^2\\
              & = \frac{16}{25} - \frac{9}{25}\\
              & = \frac{7}{25}\\
\tan(2\theta) & = \frac{\sin(2\theta)}{\cos(2\theta)}\\
              & = \frac{\frac{24}{25}}{\frac{7}{25}}\\
              & = \frac{24}{7}
\end{align*}
as you found.  
