If tan A and tan B are the roots of the quadratic equation, 3 x^2-10 x-25=0 then the value of 3 sin ^2(A+B)-10 sin (A+B) · cos (A+B)-25 cos ^2 (A+B)…

As $\tan A$tanA and $\tan B$tanB are the roots of $3 x^2-10 x-25=0$3x2−10x−25=0 So, $\tan (A+B)=\frac{\tan A+\tan B}{1-\tan A \tan B}=\frac{\frac{10}{3}}{1+\frac{25}{3}}$tan(A+B)=1−tanAtanBtanA+tanB​=1+325​310​​ $=\frac{10 / 3}{28 / 3}=\frac{5}{14}$=28/310/3​=145​ Now, $\cos ^2(A+B)=-1+2 \cos ^2(A+B)$cos2(A+B)=−1+2cos2(A+B) $=\frac{1-\tan ^2(A+B)}{1+\tan ^2(A+B)} \Rightarrow \cos ^2(A+B)=\frac{196}{221}$=1+tan2(A+B)1−tan2(A+B)​⇒cos2(A+B)=221196​ $\therefore 3 \sin ^2(A+B)-10 \sin (A+B)$∴3sin2(A+B)−10sin(A+B) $=\cos ^2(A+B)\left[3 \tan ^2(A+B)-10 \tan (A+B)-25\right]$=cos2(A+B)[3tan2(A+B)−10tan(A+B)−25] $=\frac{75-700-4900}{196} \times \frac{196}{221}=-\frac{5525}{196} \times \frac{196}{221}=-25$=19675−700−4900​×221196​=−1965525​×221196​=−25