Q. 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) $2n\pi,\ 2n\pi + \frac{\pi}{4}$
C) $2n\pi$
D) None of these
Explanation

(b)
Given, $1 - \cos{x = \left( \sqrt{2} - 1 \right)\sin x}$
$\Rightarrow \ 2\sin{\frac{x}{2}\left( \sin{\frac{x}{2} - (\sqrt{2} - 1)}\cos\frac{x}{2} \right) = 0}$
$\Rightarrow \ \sin{\frac{x}{2} = 0}$ or $\tan{\frac{x}{2} = \sqrt{2} - 1 = \tan{\frac{45{^\circ}}{2}\ }}$
$\Rightarrow \ \frac{x}{2} = n\pi,\ \ \frac{x}{2} = n\pi + \frac{\pi}{8}$
$\Rightarrow \ \ x = 2n\pi,\ \ 2n\pi + \frac{\pi}{4}$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.