Solution

We write \(tan^{-1}(\frac{cos x}{1-sin x}) = tan^{-1}[\frac{cos^2(\frac{x}{2}) - sin^2(\frac{x}{2})}{cos^2(\frac{x}{2}) + sin^2(\frac{x}{2}) - 2sin(\frac{x}{2})cos(\frac{x}{2})}]\) \(= tan^{-1}[\frac{(cos(\frac{x}{2}) + sin(\frac{x}{2}))(cos(\frac{x}{2}) - sin(\frac{x}{2}))}{(cos(\frac{x}{2}) - sin(\frac{x}{2}))^2}]\) \(= tan^{-1}[\frac{cos(\frac{x}{2}) + sin(\frac{x}{2})}{cos(\frac{x}{2}) - sin(\frac{x}{2})}] = tan^{-1}[\frac{1+tan(\frac{x}{2})}{1-tan(\frac{x}{2})}]\) \(= tan^{-1}[tan(\frac{\pi}{4} + \frac{x}{2})] = \frac{\pi}{4} + \frac{x}{2}\)