As #tanu=(2tan(u/2))/(1-tan^2(u/2)#
#(2tan(u/2))/((1-tan^2(u/2)))=-5/8#
or #16tan(u/2))=-5+5tan^2(u/2)#
or #5tan^2(u/2)-16tan(u/2)-5=0#
and using quadratic formula
#tan(u/2)=(-(-16)+-sqrt((-16)^2-4xx5xx(-5)))/10#
= #(16+-sqrt(256+100))/10=(16+-sqrt356)/10=(8+-sqrt89)/5#
As #(3pi)/2 < u < 2pi#, we have #(3pi)/4 < u/2 < pi# i.e. #u/2# lies in Q2
and hence #tan(u/2)# is negative i.e. #tan(u/2)=-(sqrt89-8)/5#
and #sec(u/2)=-sqrt(1+(sqrt89-8)^2/25)#
= #-sqrt(1+(89+64-16sqrt89)/25)#
= #-sqrt((178-16sqrt89)/25)#
and #cos(u/2)=-5/sqrt((178-16sqrt89)#
and #sin(u/2)=cos(u/2)tan(u/2)=-5/sqrt((178-16sqrt89))xx(-(sqrt89-8)/5)#
= #(sqrt89-8)/sqrt(178-16sqrt89)#