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: \(\boxed{-0.4472135954999579}\)