Question

The most general solutions of the equation $\sec x-1=(\sqrt{2}-1)\tan x$ are given by

• A. $n\pi +\frac{\pi }{8}$
• B. $n\pi \pm \frac{\pi }{4}$
• C. $n\pi \pm \frac{\pi }{8}$
• D. none of these

Answer 1

The correct option is B $n\pi \pm \frac{\pi }{4}$

$\sec x=(\sqrt{2}-1)\tan x+1$
$1+{\tan }^{2}x=(3-2\sqrt{2}){\tan }^{{}^2}x+1+2\tan x(\sqrt{2}-1)$
${\tan }^{2}x(2-2\sqrt{2})+2\tan x(\sqrt{2}-1)=0$
${\tan }^{2}x(1-\sqrt{2})+\tan x(\sqrt{2}-1)=0$
$\tan x\left(\tan x\right(1-\sqrt{2})+(\sqrt{2}-1)=0$
$\tan x=1$
$x=n\pi \pm \frac{\pi }{4}$