Okay. We've got:
#costheta/(1-sintheta)+costheta/(1+sintheta)=4#
Let's ignore the #RHS# for now.
#costheta/(1-sintheta)+costheta/(1+sintheta)#
#(costheta(1+sintheta)+costheta(1-sintheta))/((1-sintheta)(1+sintheta))#
#(costheta((1-sintheta)+(1+sintheta)))/(1-sin^2theta)#
#(costheta(1-sintheta+1+sintheta))/(1-sin^2theta)#
#(2costheta)/(1-sin^2theta)#
According to the Pythagorean Identity,
#sin^2theta+cos^2theta=1#. So:
#cos^2theta=1-sin^2theta#
Now that we know that, we can write:
#(2costheta)/cos^2theta#
#2/costheta=4#
#costheta/2=1/4#
#costheta=1/2#
#theta=cos^-1(1/2)#
#theta=pi/3#, when #0<=theta<=pi#.
In degrees, #theta=60^@# when #0^@<=theta<=180^@#