(tan A +sec A - 1) / (tan A - sec A+ 1)= (1+sin A)/ cos A

Can anyone prove this identity for me? I will really appreciate it.

Use
sec^2A-tan^2A=1
take the numerator
secA+tanA-1=
secA+tanA-(sec^2A-tan^2A)=(secA+tanA)-…
now take (secA+tanA) as common
=(secA+tanA)(1-secA+tanA)
NOW Nr/Dr gives
secA+tanA
=(1/cosA)+(sinA/cosA)=(1+sinA)/cosA

(tan A +sec A - 1) / (tan A - sec A+ 1)
= (tan A +sec A - (sec^2 A- tan ^2 A)) / (tan A - sec A+ 1)
= ((tan A + sec A) - (tan A + sec A)(tan A - sec A))/(tan A - sec A +1)
= (tan A + sec A) (1+ tan A - sec A)/(1+ tan A - sec A)
= tan A + sec A
= sin A /cos A + 1/cos A
= (sin A + 1)/cos A
= RHS

substitute A=45 degrees and u can find if it correct or not :D