Question

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

• A. $-10$
• B. $10$
• C. $-25$
• D. $25$

Answer 1

The correct option is C $-25$

$3x^2-10x-25=0\Rightarrow 3x^2-15x+5x-25=0\Rightarrow (3x+5)(x-5)=0\Rightarrow x=-\frac{5}{3},5$
Now,
$\tan (A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}=\frac{5-\frac{5}{3}}{1+\frac{25}{3}}=\frac{5}{14}\cos (A+B)=\frac{14}{\sqrt{221}}$
Now,
$S=3{\sin }^2(A+B)-10\sin (A+B)\cdot \cos (A+B)-25{\cos }^2(A+B)\Rightarrow S={\cos }^2(A+B)\left[3{\tan }^2\right(A+B)-10\tan (A+B)-25]\Rightarrow S=\frac{196}{221}[3\times \frac{25}{196}-10\times \frac{5}{14}-25]\Rightarrow S=\frac{25}{221}[3-28-196]\Rightarrow S=-25$