Mathematics - Trigonometric Equation Question with Solution | TestHub

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