tanx=2 x∈（π,3π／2）,求[sin(π-α)+2sin(3π／2+α)]／[cos(3π-α)+1]的值.

[sin(π-α)+2sin(3π／2+α)]／[cos(3π-α)+1]=[sinα-2cosα]／[-cosα+1]=[sinα/cosα-2cosα/cosα]／[-cosα/cosα+1/cosα]=[tanα-2]／[-1+1/cosα]=[2-2]／[-1+1/cosα]=0／[-1+1/cosα]=0