Answer 1

We have to prove (sec A + tan A − 1) (sec A − tan A + 1) = 2 tan A

We know that `sec^2 theta A - tan^2 theta A = 1`

So, we have

(sec A + tan A - 1)(sec A - tan A +  1) = {sec A + (tan A - 1)}{sec A - (tan A - 1)}

`= sec^2 A - (tan A - 1)^2`

`= sec^2 A - (tan^2 A - 2 tan A + 1)`

`= (sec^2 A - tan^2 A) + 2 tan A - 1`

So we have

(sec A + tan A  - 1)(sec A - tan A + 1) = 1 +  tan A - 1

= 2 tan A

Hence proved.