y=1/1-tan2x；
∴1-tan2x≠0
且2x≠π/2+kπ
∴tan2x≠1且x≠π/4+kπ/2
∴2x≠π/4+kπ且x≠π/4+kπ/2
∴定义域为{x|x≠π/8+kπ/2且x≠π/4+kπ/2}tan2x≠1且x≠π/4+kπ/2这步是怎么来的啊首先分母不能等于0那么1-tan2x≠0tan2x≠0其次函数f(t)=tant中定义域为t≠π/2+kπ∴2x≠π/2+kπ，即x≠π/4+kπ/2所以要同时满足tan2x≠1且x≠π/4+kπ/2