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

Answer: (C)

$3x^2-10x-25=0$
$\tan A+\tan B=\frac{10}{3}$
$\tan A+\tan B=-\frac{23}{3}$
$\tan (A+B)=\frac{\tan A+\tan }{1}$
$=\frac{\frac{10}{3}}{1+\frac{23}{3}}$
$=\frac{10}{28}=\frac{5}{14}$
Divide and multiply by ${\cos }^2\times (A+B)$
$3{\tan }^2(A+B)-10\tan (A+B)-25({\cos }^2(A+B)$
$3\frac{25}{196}-10\left(\frac{5}{14}\right)-25\left({\cos }^2(A+B)\right)$
$\frac{75-700-4500}{196}\left({\cos }^2(A+B)\right)$
$-\frac{5525}{196}\left(\frac{1}{1+{\tan }^2(A+B)}\right)$
$-\frac{5525}{196}\left(\frac{1}{1+\frac{25}{196}}\right)$
$=\frac{-5521}{221}$