#theta /4 in Q_1 and theta/2 in Q_2#,

for #0 < theta in Q_4#, and then, cos (theta/2) < 0#.

#f(theta) = 3cos (theta/2)/sin(theta/2) + sin(theta/2)/cos(theta/2)-3 ( sqrt (1/2(1 + cos(theta/2))#

#= (3 cos^2(theta/2) + sin ^2(theta/2))/(sin(theta/2)cos(theta/2))#

#-3 sqrt (1/2(1 + cos(theta/2))#

#=( 3 ( 1/2 ( 1 + cos theta ))+1/2( 1 - cos theta ))/(1/2 sin theta )#

#-3 (sqrt (1/2(1 +- sqrt(1/2(1+ cos theta ))))#

#= 2(2csc theta + cot theta )-(3/8)sqrt(8sqrt8) (sqrt ((2sqrt2 +- sqrt(1+ cos theta ))))#,

choosing #-# of #(+-)#, for # 0 < theta in Q_4#.

Note that,

if #0 < theta in Q_4 in ( - pi/2, 0 ), theta/2 in Q_4 in ( - pi/4, 0 )#

and, the choice is + ( for #cos ( theta/2 )#). Indeed, incredible.

I am not approving ( like my invalidating #r < 0# ), use of negative

#theta#. This contradiction is the reason.

As these are not in tune with the universal faith on conventions, I

do not invite any discussion. I would like to be alone as non-

conventional, on such foundation matters.