As tanA and tanB are the roots of 3x2−10x−25=0 So, tan(A+B)=1−tanAtanBtanA+tanB=1+325310 =28/310/3=145 Now, cos2(A+B)=−1+2cos2(A+B) =1+tan2(A+B)1−tan2(A+B)⇒cos2(A+B)=221196 ∴3sin2(A+B)−10sin(A+B) =cos2(A+B)[3tan2(A+B)−10tan(A+B)−25] =19675−700−4900×221196=−1965525×221196=−25