3.323. $\sin \alpha \sin ^{2}\left(\alpha-270^{\circ}\right)\left(1+\operatorname{tg}^{2} \alpha\right)+\cos \alpha \cos ^{2}\left(\alpha+270^{\circ}\right)\left(1+\operatorname{ctg}^{2} \alpha\right)$.
## Solution.

$$
\begin{aligned}
& \sin \alpha \sin ^{2}\left(\alpha-270^{\circ}\right)\left(1+\operatorname{tg}^{2} \alpha\right)+\cos \alpha \cos ^{2}\left(\alpha+270^{\circ}\right)\left(1+\operatorname{ctg}^{2} \alpha\right)= \\
& =\sin \alpha\left(\sin \left(270^{\circ}-\alpha\right)\right)^{2}\left(1+\operatorname{tg}^{2} \alpha\right)+\cos \alpha\left(\cos \left(270^{\circ}+\alpha\right)\right)^{2}\left(1+\operatorname{ctg}^{2} \alpha\right)= \\
& =\sin \alpha \cos ^{2} \alpha\left(1+\operatorname{tg}^{2} \alpha\right)+\cos \alpha \sin ^{2} \alpha\left(1+\operatorname{ctg}^{2} \alpha\right)=
\end{aligned}
$$

$=\sin \alpha \cos ^{2} \alpha\left(1+\frac{\sin ^{2} \alpha}{\cos ^{2} \alpha}\right)+\cos \alpha \sin ^{2} \alpha\left(1+\frac{\cos ^{2} \alpha}{\sin ^{2} \alpha}\right)=$

$=\frac{\sin \alpha \cos ^{2} \alpha\left(\cos ^{2} \alpha+\sin ^{2} \alpha\right)}{\cos ^{2} \alpha}+\frac{\cos \alpha \sin ^{2} \alpha\left(\sin ^{2} \alpha+\cos ^{2} \alpha\right)}{\sin ^{2} \alpha}=$

$=\sin \alpha+\cos \alpha=\frac{2}{\sqrt{2}}\left(\frac{\sqrt{2}}{2} \sin \alpha+\frac{\sqrt{2}}{2} \cos \alpha\right)=$

$=\sqrt{2}\left(\sin 45^{\circ} \cos \alpha+\cos 45^{\circ} \sin \alpha\right)=$

$=[\sin x \cos y+\cos x \sin y=\sin (x+y)]=\sqrt{2} \sin \left(45^{\circ}+\alpha\right)$.

Answer: $\sqrt{2} \sin \left(45^{\circ}+\alpha\right)$.