Value of Sine Expression Involving Roots of Quadratic Equation

Given that $\tan A$ and $\tan B$ are the roots of the quadratic equation $x^2 - 2x - 5 = 0$, where $A, B \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$, we can use Vieta's formulas to write: $\tan A + \tan B = 2$ $\tan A \cdot \tan B = -5$

Using the compound angle formula for tangent: $\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} = \frac{2}{1 - (-5)} = \frac{2}{6} = \frac{1}{3}$

To determine the quadrant of $A+B$, we solve for the roots of $x^2 - 2x - 5 = 0$: $x = \frac{2 \pm \sqrt{4 - 4(1)(-5)}}{2} = 1 \pm \sqrt{6}$

Without loss of generality, let $\tan A = 1 + \sqrt{6}$ and $\tan B = 1 - \sqrt{6}$. Since $\tan A \approx 3.449 > 1$ and $A \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$, we have $A \in \left(\frac{\pi}{4}, \frac{\pi}{2}\right)$. Similarly, since $\tan B \approx -1.449 < -1$ and $B \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$, we have $B \in \left(-\frac{\pi}{2}, -\frac{\pi}{4}\right)$.

Notice that $\tan A = 1 + \sqrt{6} > \sqrt{6} - 1 = -\tan B = \tan(-B)$. Since $A, -B \in \left(0, \frac{\pi}{2}\right)$ and the tangent function is strictly increasing on this interval, $A > -B \implies A + B > 0$. Also, adding the bounds of $A$ and $B$: $\frac{\pi}{4} - \frac{\pi}{2} < A + B < \frac{\pi}{2} - \frac{\pi}{4} \implies -\frac{\pi}{4} < A + B < \frac{\pi}{4}$

Since $A + B > 0$ and $A + B \in \left(-\frac{\pi}{4}, \frac{\pi}{4}\right)$, it follows that $A + B \in \left(0, \frac{\pi}{4}\right)$. Thus, $\cos(A+B) > 0$.

Using the identity $\cos(A+B) = \frac{1}{\sqrt{1 + \tan^2(A+B)}}$: $\cos(A+B) = \frac{1}{\sqrt{1 + \left(\frac{1}{3}\right)^2}} = \frac{1}{\sqrt{\frac{10}{9}}} = \frac{3}{\sqrt{10}}$

Now, using the half-angle formula $2\sin^2\left(\frac{\theta}{2}\right) = 1 - \cos\theta$: $20 \sin^2 \left(\frac{A + B}{2}\right) = 10 \cdot \left[ 2 \sin^2 \left(\frac{A + B}{2}\right) \right] = 10 \left(1 - \cos(A+B)\right)$

Substituting the value of $\cos(A+B)$: $20 \sin^2 \left(\frac{A + B}{2}\right) = 10 \left(1 - \frac{3}{\sqrt{10}}\right) = 10 - \frac{30}{\sqrt{10}} = 10 - 3\sqrt{10}$

Hence, the correct option is C.