一道数学题如图.
如图.
解：[sin(2π-α）tan(π+α）cos(π/2-α)]/[cos(π-α）tan(3π-α)tan(-α-π）]=[-sinα*tanα*sinα]/[-cosα(-tanα)(-tan(α+π）] =[-sinαtanαsinα]/[cosαtanα(-tanα)] =sinαsinα/cosαtanα=sinα