Prove that `(1+sectheta-tantheta)/(1+sectheta+tantheta) = (1-sintheta)/(costheta)`

Explanation

LHS = `(1+sectheta-tantheta)/(1+sectheta+tantheta)`
`(1+1//costheta-sintheta//costheta)/(1+1//costheta+sintheta//costheta)` `[therefore sectheta=1/costheta` and `tantheta=(sintheta)/(costheta)]`
`(costheta+1-sintheta)/(costheta+1+sintheta) = ((costheta+1)-sintheta)/((costheta+1)+sintheta) = (2cos^(2)theta/2. costheta/2)/(2cos^(2)theta/2+2sintheta/2.costheta/2)`
`[therefore 1+costheta=2cos^(2)theta/2` and `sintheta=2sintheta/2costheta/2]`
`=(2cos^(2)theta/2-2sintheta/2.costheta/2)/(2cos^(2)theta/2+2sintheta/2.costheta/2) = (2costheta/2(costheta/2-sintheta/2))/(2costheta/2(costheta/2+sintheta/2))`
`=(costheta/2-sintheta/2)/(costheta/2+sintheta/2) xx (costheta/2-sintheta/2)/(costheta/2-sintheta/2)` [by rationalisation]
`=(costheta/2-sintheta/2)^(2)/(cos^(2)theta/2-sin^(2)theta/2)` `[therefore (a-b)^(2)= a^(2)+b^(2)-2ab` and `(a-b)(a+b)=(a^(2)-b^(2))]`
`=(cos^(2)theta/2) + sin^(2)theta/2-(2sintheta/2.costheta/2)/(costheta)` `[therefore cos^(2)theta/2-sin^(2)theta/2=costheta]`
`(1-sintheta)/(costheta)` `[therefore sin^(2)theta/2 + cos^(2)theta/2=1]`
`(1-sintheta)/(costheta)` `[therefore sin^(@)theta/2 + cos^(2)theta/2=1]`
= RHS Hence Proved.