If we make #arctan((3pi)/2)=theta# then
#tan theta=(3pi)/2# => #sin theta/cos theta =(3pi)/2# => #sqrt(1-cos^2 theta)=(3pi)/2*cos theta# =>#1-cos^2 theta=(9pi^2)/4# => #cos^2 theta*(9pi^2+4)/4=1# => #cos theta=2/sqrt(9pi^2+4)# => #theta = arccos (2/sqrt(9pi^2+4))#
If we make #arcsec(pi/4)=phi# then
#sec phi =pi/4# => #cos phi=4/pi#
#sin phi=sqrt(1-cos^2 phi)=sqrt(1-16/pi^2)# => #sin phi=sqrt(pi^2-16)/pi# => #phi=arcsin(sqrt(pi^2-16)/pi)#
Using the results in the original expression
#cos theta-2sin phi=#
#cos (arccos(2/sqrt(9pi^2+4)))-2*sin (arcsin(sqrt(pi^2-16)/pi))=#
#=2/sqrt(9pi^2+4)-2*sqrt(pi^2-16)/pi#
