Let $y = \tan^{-1}\left(\frac{\cos x}{1-\sin x}\right)$.First, use half-angle identities to rewrite the numerator and denominator:$$\cos x = \cos^2\frac{x}{2} - \sin^2\frac{x}{2} = \left(\cos\frac{x}{2} - \sin\frac{x}{2}\right)\left(\cos\frac{x}{2} + \sin\frac{x}{2}\right)$$$$1 - \sin x = \cos^2\frac{x}{2} + \sin^2\frac{x}{2} - 2\sin\frac{x}{2}\cos\frac{x}{2} = \left(\cos\frac{x}{2} - \sin\frac{x}{2}\right)^2$$Substitute these back into the expression:$$\frac{\cos x}{1-\sin x} = \frac{\left(\cos\frac{x}{2} - \sin\frac{x}{2}\right)\left(\cos\frac{x}{2} + \sin\frac{x}{2}\right)}{\left(\cos\frac{x}{2} - \sin\frac{x}{2}\right)^2}$$Cancel the common term $\left(\cos\frac{x}{2} - \sin\frac{x}{2}\right)$:$$= \frac{\cos\frac{x}{2} + \sin\frac{x}{2}}{\cos\frac{x}{2} - \sin\frac{x}{2}}$$Divide the numerator and denominator by $\cos\frac{x}{2}$:$$= \frac{1 + \tan\frac{x}{2}}{1 - \tan\frac{x}{2}}$$Since $\tan\frac{\pi}{4} = 1$, we can rewrite this as:$$= \frac{\tan\frac{\pi}{4} + \tan\frac{x}{2}}{1 - \tan\frac{\pi}{4}\tan\frac{x}{2}}$$Using the identity $\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$:$$= \tan\left(\frac{\pi}{4} + \frac{x}{2}\right)$$Therefore:$$y = \tan^{-1}\left(\tan\left(\frac{\pi}{4} + \frac{x}{2}\right)\right)$$Since $x \in (-\frac{\pi}{2}, \frac{\pi}{2})$, the angle lies within the principal range, so:$$y = \frac{\pi}{4} + \frac{x}{2}$$