As #csctheta=-3/2# and #tantheta>0#
we have #sintheta=1/(-3/2)=-2/3#
As sine ratio is negative and tangent ratio is positive, #theta# lies in Q3 and hence #costheta# will be negative and as #pi < theta < (3pi)/2#, we have
#pi/2 < theta/2 < (3pi)/4# and #cos(theta/2)# is too negative.
and #costheta=sqrt(1-(-2/3)^2)=sqrt(1-4/9)=-sqrt5/9=-sqrt5/3#
and as #costheta=2cos^2(theta/2)-1#
#cos(theta/2)=sqrt((1+costheta)/2)==sqrt((1-sqrt5/3)/2)#
= #-sqrt((3-sqrt5)/6)=-sqrt((18-6sqrt5)/36)#
= #-sqrt((15+3-2sqrt45)/36)#
= #-sqrt(((sqrt15)^2+(sqrt3)^2-2sqrt(15xx3))/36)#
= #-(sqrt15-sqrt3)/6#
