Question

[tan(π+a)cos(2π+a)sin(a-3π/2)]/[cos(-a-3π)sin(-3π-a)]

Answer 1

原式=tanacosacosa/[(-cosa)sina]
=-tanacosa/sina
=-(sina/cosa)(cosa/sina)
=-1

Answer 2

tan(π+a)cos(2π+a)sin(a-3π/2)]/[cos(-a-3π)sin(-3π-a)]
=-tanacosa（-cosa）/（-cosa）sina=-1