Find the exact value of the function.
$\tan \frac{\beta }{2}$, given $\tan \beta =\frac{\sqrt{5}}{2}$, with ${180}^{\circ }<\beta <{270}^{\circ }$
$\tan \frac{\beta }{2}=$
(Type an exact answer, using radicals as needed. Rationalize all denominators.)

Step 1 ：We are given that $\tan \beta =\frac{\sqrt{5}}{2}$ and we need to find the exact value of $\tan \frac{\beta }{2}$.

Step 2 ：We know that the formula for $\tan \frac{\beta }{2}$ in terms of $\tan \beta$ is given by: $\tan \frac{\beta }{2}=\frac{1-\cos \beta }{\sin \beta }$.

Step 3 ：We can express $\sin \beta$ and $\cos \beta$ in terms of $\tan \beta$ as follows: $\sin \beta =\frac{\tan \beta }{\sqrt{1+{\tan }^2\beta }}$ and $\cos \beta =\frac{1}{\sqrt{1+{\tan }^2\beta }}$.

Step 4 ：Substituting these into the formula for $\tan \frac{\beta }{2}$, we get: $\tan \frac{\beta }{2}=\frac{1-\frac{1}{\sqrt{1+{\tan }^2\beta }}}{\frac{\tan \beta }{\sqrt{1+{\tan }^2\beta }}}$.

Step 5 ：We can simplify this to: $\tan \frac{\beta }{2}=\frac{\sqrt{1+{\tan }^2\beta }-1}{\tan \beta }$.

Step 6 ：Substituting the given value of $\tan \beta =\frac{\sqrt{5}}{2}$ into this equation, we find the exact value of $\tan \frac{\beta }{2}$ to be approximately 0.4472135954999579.

Step 7 ：However, we need to consider the quadrant of $\beta$ to determine the sign of $\tan \frac{\beta }{2}$. Since ${180}^{\circ }<\beta <{270}^{\circ }$, $\beta$ is in the third quadrant where both sine and cosine are negative. Therefore, $\tan \frac{\beta }{2}$ should be negative.

Step 8 ：So, the exact value of $\tan \frac{\beta }{2}$ is $-0.4472135954999579$.

Step 9 ：Final Answer: $\boxed{{\displaystyle -0.4472135954999579}}$