f(x)=cosx√((1-sinx)/(1+sinx))+sinx√((1-cosx)/(1+cosx))
=√[(1-sinx)(1-sin^2x)/(1+sinx)]+√[(1-cosx)*(1-cos^2x)/(1+cosx)]
=√[(1-sinx)^2(1+sinx)/(1+sinx)]+√[(1-cosx)^2*(1+cosx)/(1+cosx)]
=√(1-sinx)^2+√(1-cosx)^2
=1-sinx+1-cosx
=2-sinx-cosx
=2-2(√2/2sinx+√2/2cosx)
=2-2sin(x+π/4)
f（π/4）=2-2sin(π/4+π/4)=2-2sinπ/2=2-2=0
函数在（π/2,π）是减函数
-1