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

Explanation

Note : In such type of questions ,it is better to write `sec^(2)theta-tan^(2)thetaorcosec^(2)theta-cot^(2)theta`for1 in only numberator . If in R.H.S. the single term in either number numerator of denominator is `sin theta` then convert the question in cosec `theta and cotthetaif` single term is `costheta` then convert the question in `secthetaand tantheta`. As in this question in R.H.S . single term `sintheta` is in numerator so we will use `"cosec"^(2)theta-cot^(2)theta`for 1.
L.H.S.`=(sectheta+1-tantheta)/(tantheta+1-sectheta)=(1/costheta+1-(sintheta)/(costheta))/(sintheta/(costheta)+1-(1)/(costheta))`
`=(1+costheta-sintheta)/(sintheta+costheta-1)=((1)/(costheta)+(costheta)/(sintheta)-(sintheta)/(sinthe))/(sintheta/(sintheta)+(costheta)/(sintheta)-(1)/(sintheta))`
(dividing Nr . and Dr . by `sintheta`to convert it in `cosecthetaandcottheta`)
`=("cosec"theta+cottheta-1)/(1+cottheta-"cosec"theta)=("cosec"theta+cottheta-("cosec"^(2)theta-cot^(2)theta))/(1+cottheta-"cosec"theta)`
`=(("cosec"theta+cottheta)-("cosec"theta+cottheta)("cosec"theta-cottheta))/(1+cottheta-"cosec"theta)`
`=(("cosec"theta+cottheta)[1-"cosec"theta+cottheta])/(1+cottheta-"cosec"theta)="cosec"theta+cottheta`
`=(1)/(sintheta)+(costheta)/(sintheta)=((1+costheta))/(sintheta)=((1+costheta)(1-costheta))/(sintheta(1-costheta))`
`=(1-cos^(2)theta)/(sintheta(1-costheta))=(sin^(2)theta)/(sintheta(1-costheta))=(sintheta)/(1-cos)=R.H.S.`