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If $\tan A$ and $\tan B$ are the roots of the quadratic equation, $3 x^2-10 x-25=0$ then the value of $3 \sin ^2(A+B)-10 \sin (A+B) \cdot \cos (A+B)-25 \cos ^2$ $(A+B)$ is
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The correct answer is:
$-25$
$-25$
As $\tan A$ and $\tan B$ are the roots of $3 x^2-10 x-25=0$
So, $\tan (A+B)=\frac{\tan A+\tan B}{1-\tan A \tan B}=\frac{\frac{10}{3}}{1+\frac{25}{3}}$ $=\frac{10 / 3}{28 / 3}=\frac{5}{14}$
Now, $\cos ^2(A+B)=-1+2 \cos ^2(A+B)$
$=\frac{1-\tan ^2(A+B)}{1+\tan ^2(A+B)} \Rightarrow \cos ^2(A+B)=\frac{196}{221}$
$\therefore 3 \sin ^2(A+B)-10 \sin (A+B)$
$=\cos ^2(A+B)\left[3 \tan ^2(A+B)-10 \tan (A+B)-25\right]$
$=\frac{75-700-4900}{196} \times \frac{196}{221}=-\frac{5525}{196} \times \frac{196}{221}=-25$
So, $\tan (A+B)=\frac{\tan A+\tan B}{1-\tan A \tan B}=\frac{\frac{10}{3}}{1+\frac{25}{3}}$ $=\frac{10 / 3}{28 / 3}=\frac{5}{14}$
Now, $\cos ^2(A+B)=-1+2 \cos ^2(A+B)$
$=\frac{1-\tan ^2(A+B)}{1+\tan ^2(A+B)} \Rightarrow \cos ^2(A+B)=\frac{196}{221}$
$\therefore 3 \sin ^2(A+B)-10 \sin (A+B)$
$=\cos ^2(A+B)\left[3 \tan ^2(A+B)-10 \tan (A+B)-25\right]$
$=\frac{75-700-4900}{196} \times \frac{196}{221}=-\frac{5525}{196} \times \frac{196}{221}=-25$
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