3.067. $\cos \alpha\left(1+\cos ^{-1} \alpha+\operatorname{tg} \alpha\right)\left(1-\cos ^{-1} \alpha+\operatorname{tg} \alpha\right)$

3.067. $\cos \alpha\left(1+\cos ^{-1} \alpha+\tan \alpha\right)\left(1-\cos ^{-1} \alpha+\tan \alpha\right)$
## Solution.

$$
\begin{aligned}
& \cos \alpha\left(1+\cos ^{-1} \alpha+\operatorname{tg} \alpha\right)\left(1-\cos ^{-1} \alpha+\operatorname{tg} \alpha\right)= \\
& =\cos \alpha\left(1+\frac{1}{\cos \alpha}+\frac{\sin \alpha}{\cos \alpha}\right)\left(1-\frac{1}{\cos \alpha}+\frac{\sin \alpha}{\cos \alpha}\right)= \\
& =\cos \alpha \cdot \frac{\cos \alpha+1+\sin \alpha}{\cos \alpha} \cdot \frac{\cos \alpha-1+\sin \alpha}{\cos \alpha}=
\end{aligned}
$$

$$
\begin{aligned}
& =\frac{((\cos \alpha+\sin \alpha)+1)((\cos \alpha+\sin \alpha)-1)}{\cos \alpha}=\frac{(\cos \alpha+\sin \alpha)^{2}-1}{\cos \alpha}= \\
& =\frac{\cos ^{2} \alpha+2 \sin \alpha \cos \alpha+\sin ^{2} \alpha-1}{\cos \alpha}=\frac{2 \sin \alpha \cos \alpha}{\cos \alpha}=2 \sin \alpha
\end{aligned}
$$

Answer: $2 \sin \alpha$.