3. a) Simplify the expression $\quad A=\frac{\sin 2 x-\cos 2 x+1}{\sin 2 x+\cos 2 x-1}$.

b) For which values of $x$, $A>0$?
Solution. a) Using trigonometric identities, we have

$$
\begin{aligned}
A & =\frac{\sin 2 x-\cos 2 x+1}{\sin 2 x+\cos 2 x-1}=\frac{2 \sin x \cos x-\cos ^{2} x+\sin ^{2} x+1}{2 \sin x \cos x+\cos ^{2} x-\sin ^{2} x+1} \\
& =\frac{2 \sin x(\cos x+\sin x)}{2 \sin x(\cos x-\sin x)}=\frac{\cos x+\sin x}{\cos x-\sin x}=\frac{1+\operatorname{tg} x}{1-\operatorname{tg} x}
\end{aligned}
$$

where $\sin x \neq 0$ and $\cos x \neq 0$.

b) We have, $A>0$ if and only if $\cos ^{2} x-\sin ^{2} x=\cos 2 x>0$. According to this,

$$
\begin{aligned}
& 2 x \in\left(2 k \pi, \frac{\pi}{2}+2 k \pi\right) \cup\left(\frac{3 \pi}{2}+2 k \pi, 2 \pi+2 k \pi\right) \\
& x \in\left(k \pi, \frac{\pi}{4}+k \pi\right) \cup\left(\frac{3 \pi}{4}+k \pi, \pi+k \pi\right), k \in \mathbb{Z}
\end{aligned}
$$