Question

Solve : $\sec x-1=(\sqrt{2}-1)\tan x,x\ne (2n-1)\frac{\pi }{2},n\in Z$

• A. $2n\pi ,(4n+1)\frac{\pi }{2},n\in Z$
• B. $n\pi ,(4n+1)\frac{\pi }{2},n\in Z$
• C. $2n\pi ,2n\pi +\frac{\pi }{4},n\in Z$
• D. $2n\pi ,(2n+1)\frac{\pi }{2},n\in Z$

Answer 1

The correct option is C $2n\pi ,2n\pi +\frac{\pi }{4},n\in Z$

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

$\frac{1}{\cos x}-1=\tan \frac{\pi }{8}\tan x$

$\frac{1-\cos x}{\cos x}=\frac{\sin x\sin \frac{\pi }{8}}{\cos x\cos \frac{\pi }{8}}$

$\cos \frac{\pi }{8}-\cos x\cos \frac{\pi }{8}=\sin x\sin \frac{\pi }{8}$

$\sin x\sin \frac{\pi }{8}+\cos x\cos \frac{\pi }{8}=\cos \frac{\pi }{8}$

$\cos (x-\frac{\pi }{8})=\cos \frac{\pi }{8}$

$x-\frac{\pi }{8}=2n\pi \pm \frac{\pi }{8},n\in Z$

$x=2n\pi +\frac{\pi }{4}$ or, $x=2n\pi ,n\in Z$