If tan A and tan B are the roots of the quadratic equation 3 x ^ 2...

Since, $tan A$ and $tan B$ are roots of the equation $3{x}^{2} - 10x - 25 = 0$.

So, $tan A + tan B = \dfrac{10}{3}$ and $tan A . tan B = - \dfrac{25}{3}$

$∴ \tan \left(A + B\right) = \dfrac{tan A + tan B}{1 - tan A . tan B} = \dfrac{\dfrac{10}{3}}{1 + \dfrac{25}{3}} = \dfrac{10}{28} = \dfrac{5}{14}$

So, $\sin \left(A + B\right) = \dfrac{\pm 5}{\sqrt{221}} \text{ and } \cos \left(A + B\right) = \dfrac{\pm 14}{\sqrt{221}}$

(Note that either both are negative or both positive)

$∴ 3\sin^{2} \left(A + B\right) - 10\sin \left(A + B\right) . \cos \left(A + B\right) - 25\cos^{2} \left(A + B\right)$

$= 3 \times \dfrac{25}{221} - \dfrac{10 \times 5 \times 14}{221} - 25 \times \dfrac{{14}^{2}}{221}$

$= \dfrac{25}{221}\left(3 - 28 - 196\right) = - 25$