Question

化简sin(2π-α）*tan(π+α）*cot(-α-π）/cos(π-α）*tan（3π-α）（急）

希望能有详细的解题过程
cot(-α-π）这部希望能说明下
请高手帮忙了~谢谢了

Answer 1

sin(2π-α)=sin(-α)=-sinα
tan(π+α)=tanα
cot(-α-π)=-cot(α+π)=-cotα
cos(π-α)=-cosα
tan(3π-α)=tan(-α)=-tanα

所以原式=sinαtanαcotα/cosαtanα
=sinα/(cosα*sinα/cosα)
=sinα/sinα
=1

Answer 2

[sin(2π－α)tan(π＋α)cot(－π－α)]/[cos(π－α)tan(3π－α)]
=-sinα*tanα(-cotα)/[(-cosα)(-tanα)]
=sinα/sinα
=1