Question

Express `tan^-1(cos x/(1-sin x)), "where" (-pi)/2 <x<pi/2` in the simplest form.

Answer 1

`tan^-1(cos x/(1-sin x))       ...[therefore cos x = cos^2  x/2 - sin^2  x/2]`

`= tan^-1 [(cos^2  x/2-sin^2  x/2)/((cos  x/2 - sin  x/2))]     ...[a^2-b^2 = (a+b)(a-b)]` 

`= tan^-1 [(cos  x/2+sin  x/2)/((cos  x/2 - sin  x/2))]`

`= tan^-1 [(1+tan  x/2)/(1 - tan  x/2)]`

`= tan^-1 [(tan  pi/4+tan  x/2)/(1 - tan  pi/4 tan  x/2)]`      ...`[tan(A+B) = (tanA+tanB)/(1-tanA.tanB)]`

`=tan^-1 (tan  pi/4+x/2)`

`=pi/4+x/2`