Solve: $\sqrt{2} \sec \theta + \tan \theta = 1$
$\sqrt{2} \sec \theta + \tan \theta = 1$
i.e. $\;$ $\dfrac{\sqrt{2}}{\cos \theta} + \dfrac{\sin \theta}{\cos \theta} = 1$
i.e. $\;$ $\sqrt{2} + \sin \theta = \cos \theta$
i.e. $\;$ $\cos \theta - \sin \theta = \sqrt{2}$
i.e. $\;$ $\dfrac{1}{\sqrt{2}} \cos \theta - \dfrac{1}{\sqrt{2}} \sin \theta = 1$
i.e. $\;$ $\cos \theta \; \cos \left(\dfrac{\pi}{4}\right) - \sin \theta \; \sin \left(\dfrac{\pi}{4}\right) = 1$
i.e. $\;$ $\cos \left(\theta + \dfrac{\pi}{4}\right) = 1$ $\;\;\;$ [i.e. $\;$ $\cos \left(\theta + \dfrac{\pi}{4}\right)$ is positive]
$\left\{\text{Principal value of } \cos x \text{ lies in } \left[0, \pi\right] \text{ i.e. in first or second quadrants} \right\}$
$\therefore \;$ We have $\cos \left(\theta + \dfrac{\pi}{4}\right) = 1 = \cos 0$
i.e. $\;$ $\theta + \dfrac{\pi}{4} = 2 n \pi \pm 0$
i.e. $\;$ $\theta = 2 n \pi - \dfrac{\pi}{4}, \;\;\; n \in Z$