1,f(x+2a)=[1+f(x+a)]/[1-f(x+a)]
={1+[1+f(x)]/[1-f(x)]}/{1-+[1+f(x)]/[1-f(x)]}
=-1/f(x)
所以f(x+4a)=f(x),即T=4a.
3,原式=
[1+2cos(x/2)平方-1+2sin(x/2)cos(x/2)]/[1+2cos(x/2)平方-1-2sin(x/2)cos(x/2)]
=[2cos(x/2)平方+2sin(x/2)cos(x/2)]/[2cos(x/2)平方-2sin(x/2)cos(x/2)]
=[cos(x/2)+sin(x/2)]/[cos(x/2)-sin(x/2)]
=[cos(x/2)+sin(x/2)]平方/[cos(x/2)平方-sin(x/2)平方]
=[1+2sin(x/2)cos(x/2)]/cosx
=(1+sinx)/cosx
不好意思,其他我也不会啊 如果你会了,记得告诉我哦!