Explanation

(b)
Given, $\sec{x - 1 = \left( \sqrt{2} - 1 \right)\tan x}$
$\Rightarrow \ \frac{1 - \cos x}{\cos x} = (\sqrt{2} - 1)\frac{\sin x}{\cos x}$
$\Rightarrow \ 2\sin^{2}{\frac{x}{2} - \left( \sqrt{2} - 1 \right)2\sin{\frac{x}{2}\cos{\frac{x}{2} = 0}}}$
$\Rightarrow \ \sin{\frac{x}{2}\left\lbrack \sin{\frac{x}{2} - (\sqrt{2} - 1)}\cos\frac{x}{2} \right\rbrack = 0}$
$\Rightarrow \ \ \sin\frac{x}{2} = 0$ or $\sin{\frac{x}{2} - \left( \sqrt{2} - 1 \right)\cos\frac{x}{2} = 0}$
$\Rightarrow \ \frac{x}{2} = n\pi$ or $\tan{\frac{x}{2} = \left( \sqrt{2} - 1 \right) = \tan\frac{45{^\circ}}{2}}$
$\Rightarrow \ \ x = 2n\pi$ or $\frac{x}{2} = \frac{45{^\circ}}{2} + n\pi$
$\Rightarrow \ \ x = 2n\pi$ or $2n\pi + \frac{\pi}{4}$