`tan^-1[(cos x)/(1+sin x)]` is equal to

Options

A. `(pi)/(4) - (x)/(2)`, for `x in (-(pi)/(2), (3pi)/(2))`
B. `(pi)/(4) -(x)/(2), " for " x in (-(pi)/(2), (pi)/(2))`
C. `(pi)/(4), (x)/(2), " for " x in ((3pi)/(2), (5pi)/(2))`
D. `(pi)/(4) -(x)/(2), " for " x in (-(3pi)/(2), (pi)/(2))`

Answer

A

Explanation

`tan^(-1) [(cos x)/(1 + sin x)] = tan^(-1) [(sin [(pi//2) -x])/(1 + cos [(pi//2) - x])]`
`= tan^(-1) [((2sin[(pi//4) - (x//2)]),(cos [(pi//4) - (x//2)]))/(2 cos^(2) [(pi//4) - (x//2)]`
`= tan^(-1) tan ((pi)/(4) - (x)/(2)) = (pi)/(4) - (x)/(2)`
`rArr -(pi)/(2) lt (pi)/(4) - (x)/(2) lt (pi)/(2)`
`rArr -(3pi)/(4) lt -(x)/(2) lt (pi)/(4)`
`rArr-(pi)/(4) lt (x)/(2) lt (3pi)/(4)`
`rArr -(pi)/(2) lt x lt (3pi)/(2)`