Simplify the following expressions:

$$
[1+(\operatorname{tg} x-\operatorname{cotg} x) \sin 2 x] \cos x+\cos 3 x
$$

$$
\frac{\operatorname{tg}^{2} x-\sin ^{2} x}{\operatorname{cotg}^{2} x-\cos ^{2} x}
$$
(1) The part in square brackets is

$$
\begin{gathered}
1+\left(\frac{\sin x}{\cos x}-\frac{\cos x}{\sin x}\right) \cdot 2 \sin x \cos x=1+2 \sin ^{2} x-2 \cos ^{2} x= \\
=3 \sin ^{2} x-\cos ^{2} x=3-4 \cos ^{2} x
\end{gathered}
$$

On the other hand,

$$
\begin{aligned}
\cos 3 x & =\cos (2 x+x)=\cos 2 x \cos x-\sin 2 x \sin x= \\
& =\left(2 \cos ^{2} x-1\right) \cos x-2 \sin ^{2} x \cos x= \\
& =\cos x\left(2 \cos ^{2} x-1-2+2 \cos ^{2} x\right)=\cos x\left(4 \cos ^{2} x-3\right)
\end{aligned}
$$

It is evident that $T_{1}=0$ for all points in its domain, i.e., if $x \neq k \pi / 2$, where $k$ is an integer.

(2) In (2), we use the fact that $\sin x=\operatorname{tg} x \cos x$ and $\cos x=\operatorname{cotg} x \sin x$ (wherever $\operatorname{tg} x$ and $\operatorname{cotg} x$ are defined). By factoring and further manipulation,

$$
T_{2}=\frac{\operatorname{tg}^{2} x}{\operatorname{cotg}^{2} x} \cdot \frac{1-\cos ^{2} x}{1-\sin ^{2} x}=\frac{\operatorname{tg}^{2} x}{(1 / \operatorname{tg} x)^{2}} \cdot \frac{\sin ^{2} x}{\cos ^{2} x}=\operatorname{tg}^{6} x
$$

László Csirmaz (Budapest, I. István g. I. o. t.)