1.

`tan^-1[(cos x)/(1+sin x)]` is equal toA. `(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» Correct Answer - A
`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)`


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