Let A, B , where A, B ∈ (-(π)/(2), (π)/(2)) , be the roots of the...

Given that $\tan A and$\tan B$are the roots of the quadratic equation$x^2 - 2x - 5 = 0$.<br><br>Sum of the roots:$\tan A + \tan B = 2$<br><br>Product of the roots:$\tan A \tan B = -5$<br><br>Using the tangent addition formula:<br><br>$\tan(A+B) = \dfrac{\tan A + \tan B}{1 - \tan A \tan B} = \dfrac{2}{1 - (-5)} = \dfrac{2}{6} = \dfrac{1}{3}$<br><br>Since$A, B \in $\left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right)$ and their product $\tan A \tan B = -5 < 0$, one angle is positive and the other is negative. This implies $A+B \in \left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right)$.

Since $\tan(A+B) = \dfrac{1}{3} > 0$, the angle $A+B$ must lie in the first quadrant, i.e., $A+B \in \left(0, \dfrac{\pi}{2}\right)$. Thus, $\cos(A+B)$ is positive.

$\cos(A+B) = \dfrac{1}{\sqrt{1 + \tan^2(A+B)}} = \dfrac{1}{\sqrt{1 + \left(\dfrac{1}{3}\right)^2}} = \dfrac{3}{\sqrt{10}}$

We need to find the value of $20\sin^2\left(\dfrac{A+B}{2}\right)$. Using the half-angle identity $\sin^2\theta = \dfrac{1 - \cos(2\theta)}{2}$:

$20\sin^2\left(\dfrac{A+B}{2}\right) = 20 \left( \dfrac{1 - \cos(A+B)}{2} \right) = 10(1 - \cos(A+B))$

Substituting the value of $\cos(A+B)$:

$10\left(1 - \dfrac{3}{\sqrt{10}}\right) = 10 - \dfrac{30}{\sqrt{10}} = 10 - 3\sqrt{10}$

Answer: $10 - 3\sqrt{10}$