Level:
Project ID:
1003076703
Source Problem:
Accepted:
1
Clonable:
1
Easy:
1
The expression \( \frac{\sin 2x}{\cos^2x } \) for $x\in(-\frac{\pi}{2}, \frac{\pi}{2})$ is equal to:
\( 2\,\mathrm{tg}\,x \)
\( \frac{\sin x}{1-\sin x} \)
\( \mathrm{tg}^2 x \)
\( \mathrm{tg}\,2x \)