若sin(π/4+x)=5/13且x∈(π/4,3π/4),则(1-tanx)/(1+tanx)=

sin(π/4+x)=5/13
cos(π/4+x)=-12/13
sinx=sin(π/4+x-π/4)
=sin(π/4+x)cos(π/4)-cos(π/4+x)sin(π/4)
=5/13*√2/2+12/13*√2/2
=17√2/26
cosx=-7√2/26
tanx=-17/7
(1-tanx)/(1+tanx)=-12/5