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(C) Let $x_1 = 	an A$ and $x_2 = 	an B$ be the roots of the equation $x^2 - 2x - 5 = 0$.From the properties of roots, $x_1 + x_2 = 2$ and $x_1 x_2 =

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$10 - 3\sqrt{10}$

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(C) Let $x_1 = 	an A$ and $x_2 = 	an B$ be the roots of the equation $x^2 - 2x - 5 = 0$.From the properties of roots, $x_1 + x_2 = 2$ and $x_1 x_2 = -5$.Using the formula for $	an(A+B)$, we have $	an(A+B) = \frac{	an A + 	an B}{1 - 	an A 	an B} = \frac{2}{1 - (-5)} = \frac{2}{6} = \frac{1}{3}$.We know that $\sin^2(\frac{A+B}{2}) = \frac{1 - \cos(A+B)}{2}$.Given $	an(A+B) = \frac{1}{3}$, we can find $\cos(A+B)$. Since $	an(A+B) > 0$, $A+B$ lies in the first or third quadrant. However, since $A, B \in (-\frac{\pi}{2}, \frac{\pi}{2})$, $A+B \in (-\pi, \pi)$. For $	an(A+B) = 1/3$, $\cos(A+B) = \frac{3}{\sqrt{1^2 + 3^2}} = \frac{3}{\sqrt{10}}$.Substituting this into the identity: $\sin^2(\frac{A+B}{2}) = \frac{1 - 3/\sqrt{10}}{2} = \frac{\sqrt{10} - 3}{2\sqrt{10}}$.Finally, $20 \sin^2(\frac{A+B}{2}) = 20 	imes \frac{\sqrt{10} - 3}{2\sqrt{10}} = \frac{10(\sqrt{10} - 3)}{\sqrt{10}} = 10 - \frac{30}{\sqrt{10}} = 10 - 3\sqrt{10}$.

Let $\tan A$ and $\tan B$, where $A, B \in (-\frac{\pi}{2}, \frac{\pi}{2})$, be the roots of the quadratic equation $x^2 - 2x - 5 = 0$. Then $20 \sin^2(\frac{A+B}{2})$ is equal to:

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