Question

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

• A. $n\pi+\frac{\pi}{8}$
• B. $2n\pi$, $2n\pi+\frac{\pi}{4}$
• C. $2n\pi$
• D. None of these

Answer 1

Answer: (B)

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