Explanation

(c)
We have, $\sqrt{2}\sec{\theta + \tan{\theta = 1}}$
$\Rightarrow \ \frac{\sqrt{2}}{\cos\theta} + \frac{\sin\theta}{\cos\theta} = 1$
$\Rightarrow \ \sin\theta - \cos\theta = - \sqrt{2}$
$\Rightarrow \frac{1}{\sqrt{2}}\sin\theta - \frac{1}{\sqrt{2}}\ \cos\theta = - 1$
$\Rightarrow \frac{1}{\sqrt{2}}\cos\theta - \frac{1}{\sqrt{2}}\sin\theta = 1$
$\Rightarrow \ \cos\frac{\pi}{4}\cos\theta - \sin\frac{\pi}{4}\sin\theta = 1$
$\Rightarrow \ \cos\left( \theta + \frac{\pi}{4} \right) = \cos\theta$
$\Rightarrow \ \theta + \frac{\pi}{4} = 2n\pi \pm 0$
$\Rightarrow \ \theta = 2n\pi - \frac{\pi}{4}\$