JEE Main 2018 — Quadratic Equation Question with Solution

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

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

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

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

(Note that either both are negative or both positive)

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

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

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