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 :

• A. -10
• B. 10
• C. -25
• D. 25

Answer 1

Answer: (C)

-25

Answer 2

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