Official solution
As tan A and tan B are the roots of 3x2 - 10x - 25 = 0,
So, tan(A + B) = \[{{\tan A + \tan B} \over {1 - \tan A\tan B}}\]
= \[{{{{10} \over 3}} \over {1 + {{25} \over 3}}}\] = \[{{10/3} \over {28/3}}\] = \[{5 \over {14}}\]
Now, cos2 (A + B) = - 1 + 2 cos2 (A + B)
= \[{{1 - {{\tan }^2}(A + B)} \over {1 + {{\tan }^2}(A + B)}}\,\] \[ \Rightarrow \] cos2(A + B) = \[{{196} \over {221}}\]
\[\therefore\,\,\,\] 3sin2(A + B) - 10sin(A + B)cos(A + B) - 25 cos2(A + B)
= cos2(A + B) [ 3tan2(A + B) - 10tan(A + B) - 25]
= \[{{75 - 700 - 4900} \over {196}} \times {{196} \over {221}}\]
= - \[{{5525} \over {196}}\] \[ \times \] \[{{196} \over {221}}\] = - 25