Упростите выражение:
[m]\frac{sin(\pi+a)cos(2\pi/2-a)tg(a-\pi/2)}{cos(\pi/2+a)cos(2\pi/2+a)tg(\pi+a)} = \frac{-sin(a)(-cos(a))(-ctg(a))}{-sin(a)(-cos(a))tg(a)} =-\frac{ctg(a)}{tg(a)} = -\frac{1}{tg^2(a)}[/m]
