3. Let $\tan \alpha$ and $\tan \beta$ be two solutions of the equation $x^{2}-3 x-3=0$. Find the value of
$$
\left|\sin ^{2}(\alpha+\beta)-3 \sin (\alpha+\beta) \cos (\alpha+\beta)-3 \cos ^{2}(\alpha+\beta)\right| .
$$
(Note: $|x|$ denotes the absolute value of $x$.)
3. Answer: 3
Solution. We have $\tan \alpha+\tan \beta=3$ and $\tan \alpha \tan \beta=-3$. Hence
$$
\tan (\alpha+\beta)=\frac{3}{1-(-3)}=\frac{3}{4} \text {. }
$$

Hence
$$
\begin{aligned}
& \left|\sin ^{2}(\alpha+\beta)-3 \sin (\alpha+\beta) \cos (\alpha+\beta)-3 \cos ^{2}(\alpha+\beta)\right| \\
= & \cos ^{2}(\alpha+\beta)\left[\left(\frac{3}{4}\right)^{2}-3\left(\frac{3}{4}\right)-3\right] \\
= & \cos ^{2}(\alpha+\beta) \times\left(-\frac{75}{16}\right),
\end{aligned}
$$
and since
$$
\tan ^{2}(\alpha+\beta)=\frac{1-\cos ^{2}(\alpha+\beta)}{\cos ^{2}(\alpha+\beta)}=\frac{1}{\cos ^{2}(\alpha+\beta)}-1=\frac{9}{16},
$$
we have $\cos ^{2}(\alpha+\beta)=\frac{16}{25}$. Thus, we have
$$
\left|\sin ^{2}(\alpha+\beta)-3 \sin (\alpha+\beta) \cos (\alpha+\beta)-3 \cos ^{2}(\alpha+\beta)\right|=|-3|=3
$$