Question

The most general solutions of the equation $secx-1=\left(\sqrt{2}-1\right)tanx$ 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 $\tan \frac{x}{2}=(\sqrt{2}-1)=\tan \frac{45^{\circ }}{2}$
$\Rightarrow x=2n\pi$ or $\frac{x}{2}=\frac{45^{\circ }}{2}+n\pi$
$\Rightarrow n=2n\pi$ or $2n\pi +\frac{\pi }{4}$