π 4 − x 2 , f o r x ∈ ( − π 2 , π 2 )

π 4 − x 2 , f o r x ∈ ( − 3 π 2 , − π 2 )

π 4 − x 2 , f o r x ∈ ( − π 2 , 3 π 2 )

π 4 − x 2 , f o r x ∈ ( 3 π 2 , 5 π 2 )

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$\tan^{- 1} [\frac{\cos x}{1 + \sin x}] =$

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\frac{π}{4} - \frac{x}{2}, f o r x \in (- \frac{π}{2}, \frac{π}{2})

\frac{π}{4} - \frac{x}{2}, f o r x \in (- \frac{3 π}{2}, - \frac{π}{2})

\frac{π}{4} - \frac{x}{2}, f o r x \in (- \frac{π}{2}, \frac{3 π}{2})

\frac{π}{4} - \frac{x}{2}, f o r x \in (\frac{3 π}{2}, \frac{5 π}{2})

answer is A.

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Detailed Solution

\tan^{- 1} [\frac{\cos x}{1 + \sin x}] = \tan^{- 1} [\frac{\sin (\frac{π}{2} - x)}{1 + \cos (\frac{π}{2} - x)}]

$= \tan^{- 1} [\frac{2 \sin (\frac{π}{4} - \frac{x}{2}) . \cos (\frac{π}{4} - \frac{x}{2})}{2 \cos^{2} (\frac{π}{4} - \frac{x}{2})}]$

$= \tan^{- 1} (\tan (\frac{π}{4} - \frac{x}{2})) = \frac{π}{4} - \frac{x}{2} i f f n \in (- \frac{π}{2}, \frac{3 π}{2})$

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