The simplified form of \(\tan^{-1}\left(\displaystyle\frac{\cos x}{1 + \sin x}\right), -\displaystyle\frac{\pi}{2} \lt x \lt \frac{\pi}{2}\) is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(\displaystyle\frac{\pi}{4} - \frac{x}{2}\)
By the half-angle formulas,
\(\displaystyle \cos x = \cos^2\frac{x}{2} - \sin^2\frac{x}{2}\)
\(\displaystyle 1 + \sin x = \left(\cos\frac{x}{2} + \sin\frac{x}{2}\right)^2\)
Cancelling the common factor \(\cos\frac{x}{2} + \sin\frac{x}{2}\), and then dividing above and below by \(\cos\frac{x}{2}\):
\(\displaystyle \frac{\cos x}{1 + \sin x} = \frac{\cos\frac{x}{2} - \sin\frac{x}{2}}{\cos\frac{x}{2} + \sin\frac{x}{2}}\)
\(\displaystyle = \frac{1 - \tan\frac{x}{2}}{1 + \tan\frac{x}{2}}\)
\(\displaystyle = \tan\left(\frac{\pi}{4} - \frac{x}{2}\right)\)
For \(-\frac{\pi}{2} \lt x \lt \frac{\pi}{2}\), the angle \(\frac{\pi}{4} - \frac{x}{2}\) lies between 0 and \(\frac{\pi}{2}\), inside the principal value range of \(\tan^{-1}\). So
\(\displaystyle \tan^{-1}\left(\frac{\cos x}{1 + \sin x}\right) = \frac{\pi}{4} - \frac{x}{2}\)
Shortcut: Put \(x = \frac{\pi}{6}\):
\(\displaystyle \tan^{-1}\left(\frac{\sqrt{3}/2}{3/2}\right) = \tan^{-1}\frac{1}{\sqrt{3}} = \frac{\pi}{6}\)
Options (1), (2), (3) and (4) give \(\frac{\pi}{3}\), \(\frac{\pi}{6}\), \(\frac{2\pi}{3}\) and \(\frac{5\pi}{12}\), so only Option (2) matches.