(i) Let \(y = \frac{1}{\cos \theta} = \sec \theta\). The derivative of \(\sec \theta\) is \(\sec \theta \tan \theta\). Thus, \(\frac{dy}{d\theta} = \sec \theta \tan \theta\).

(ii) Multiply numerator and denominator of \(\frac{1 + \sin \theta}{1 - \sin \theta}\) by \(1 + \sin \theta\) to get \(\frac{(1 + \sin \theta)^2}{(1 - \sin \theta)(1 + \sin \theta)} = \frac{1 + 2\sin \theta + \sin^2 \theta}{1 - \sin^2 \theta}\). Since \(1 - \sin^2 \theta = \cos^2 \theta\), this becomes \(\frac{1 + 2\sin \theta + \sin^2 \theta}{\cos^2 \theta}\). Using \(\sin^2 \theta = 1 - \cos^2 \theta\), we have \(\frac{2 - \cos^2 \theta + 2\sin \theta}{\cos^2 \theta} = \frac{2}{\cos^2 \theta} + \frac{2\sin \theta}{\cos^2 \theta} - 1\). This simplifies to \(2\sec^2 \theta + 2\sec \theta \tan \theta - 1\).

(iii) Using the identity from part (ii), the integral becomes \(\int_{0}^{\frac{\pi}{4}} (2\sec^2 \theta + 2\sec \theta \tan \theta - 1) \, d\theta\). This can be split into three integrals: \(2\int_{0}^{\frac{\pi}{4}} \sec^2 \theta \, d\theta + 2\int_{0}^{\frac{\pi}{4}} \sec \theta \tan \theta \, d\theta - \int_{0}^{\frac{\pi}{4}} 1 \, d\theta\). The first integral is \(2\tan \theta \bigg|_{0}^{\frac{\pi}{4}} = 2(1 - 0) = 2\). The second integral is \(2\sec \theta \bigg|_{0}^{\frac{\pi}{4}} = 2(\sqrt{2} - 1)\). The third integral is \(\theta \bigg|_{0}^{\frac{\pi}{4}} = \frac{\pi}{4}\). Combining these, the result is \(2 + 2\sqrt{2} - 2 - \frac{\pi}{4} = 2\sqrt{2} - \frac{\pi}{4}\).