f(x) = {1-√2sin(2x-π/4)}/cosx
= {1-(√2sin2xcosπ/4-cos2xsinπ/4)}/cosx
= {1-(sin2x-cos2x)}/cosx
= {1-sin2x+cos2x}/cosx
= {1-2sinxcosx+2cos^2x-1}/cosx
= {-2sinxcosx+2cos^2x}/cosx
= 2(cosx-sinx)
要使函数有意义,则必须要求sin(2x-π/4)≥0且cosx≠0,
所以要求2kπ≤2x-π/4≤2kπ+π,(k∈Z)且x≠kπ +π/2,(k∈Z)
即kπ+π/8≤x≤kπ+5π/8且x≠kπ +π/2,(k∈Z),所以f(x)的定义域为 [kπ+π/8,kπ +π/2)∪(kπ +π/2,kπ+5π/8] (k∈Z)解:(1)要使函数有意义，则必须要求sin(2x-π/4)≥0且cosx≠0，
所以要求2kπ≤2x-π/4≤2kπ+π,(k∈Z)且x≠kπ +π/2,(k∈Z)
即kπ+π/8≤x≤kπ+5π/8且x≠kπ +π/2,(k∈Z)，
所以f(x)的定义域为 [kπ+π/8,kπ +π/2)∪(kπ +π/2,kπ+5π/8] (k∈Z)


(2)f(x) = {1-√2sin(2x-π/4)}/cosx
= {1-(√2sin2xcosπ/4-cos2xsinπ/4)}/cosx
= {1-(sin2x-cos2x)}/cosx
= {1-sin2x+cos2x}/cosx
= {1-2sinxcosx+2cos^2x-1}/cosx
= {-2sinxcosx+2cos^2x}/cosx
= 2(cosx-sinx)为什么sin(2x-∏/4)≥0不好意思啊，第一题我修正一下:
（1）∵依题意，有cosx≠0
∴解得x≠kp+ π\x09/2 ，
∴f（x）的定义域为{x|x∈R，且x≠kp+ π\x09/2 ，k∈Z刚开始打字的时候把式子弄反了，第二小问发现了，修正了，忘了修正第一小问了。抱歉希望可以帮到你，对你有帮助，还望您认可一下我的解答！