Question

Solve $\sec x-1=(\sqrt{2}-1)\tan x$

Answer 1

Given, $\sec x-1=\left(\sqrt{2-1}\right)\tan x\Rightarrow -1+\sec x+\tan x-\sqrt{2}\tan x=0$
Put $y=\tan \frac{x}{2}$
$\frac{1}{y^2-1}-\frac{2y}{y^2-1}+\frac{2\sqrt{2}y}{y^2-1}-\frac{y^2}{y^2-1}+\frac{1}{\frac{1}{y^2+1}-\frac{y^2}{y^2+1}}=0$
$\Rightarrow \frac{y^2-\sqrt{2y}+y}{y^2-1}=0\Rightarrow y^2-\sqrt{2y}+y=0\Rightarrow y(-y-1+\sqrt{2})=0$
$\Rightarrow y=0$ or $y=\sqrt{2}-1$
$\Rightarrow \tan \frac{x}{2}=0$ or $\tan \frac{x}{2}=\sqrt{2}-1$
$\Rightarrow x=2n\pi$ or $x=2n\pi +\frac{\pi }{4}$