Question

pls solve this question as quickly as possible

Answer 1

Please read the full Answer:

Taking first expression

secθ+1-tanθ/secθ+1+tanθ

=(1/secθ+tanθ)+1/secθ+1+tanθ

=secθ+tanθ+1/(secθ+tanθ)(secθ+1+tanθ)

=1/secθ+tanθ

=secθ-tanθ -------------------(1)

Taking second expression

secθ-1+tanθ/1+tanθ-secθ

=(1/secθ-tanθ)-1/1+tanθ-secθ

=1+tanθ-secθ/(secθ-tanθ)(1+tanθ-secθ)

=1/secθ-tanθ

=secθ+tanθ---------------------(2)

adding (1) and (2),

secθ-tanθ+secθ+tanθ

=2secθ

Hence Proved.

Note:- I replaced secθ+tanθ by 1/secθ-tanθ

and I replaced secθ-tanθ by 1/secθ+tanθ because

sec^2θ -tan^2θ =1

so (secθ+tanθ)(secθ-tanθ)=1

so secθ+tanθ=1/secθ-tanθ

and secθ-tanθ=1/secθ+tanθ