Question

If $\tan A$ and $\tan B$ are the roots of the quadratic equation, $3x^2-10x-25=0$, then the value of $3{\sin }^2(A+B)-10\sin (A+B)\cos (A+B)-25{\cos }^2(A+B)$ is:

Answer 1

Given, $3x^2-10x-25=0$

$\tan A+\tan B=\frac{10}{3}$

$\tan A\times tanB=\frac{-25}{3}$

$\tan (A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}$

$\frac{\frac{10}{3}}{1+\frac{25}{3}}$

$\frac{\frac{10}{3}}{\frac{28}{3}}=\frac{10}{28}=\frac{5}{14}$

$\therefore \tan (A+B)=\frac{5}{14}$

$\therefore \sin (A+B)=\frac{5}{\sqrt{221}}$

$\cos (A+B)=\frac{14}{\sqrt{221}}$

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

$=3\times \frac{25}{221}-10\times \frac{70}{221}-25\times \frac{196}{221}$

$=\frac{75-700-4900}{221}$

$=-25$