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 :
$$2n\pi, 2n\pi + \frac{\pi }{4}$$
Solution :
$$\eqalign{
& 1 - \cos x = \left( {\sqrt 2 - 1} \right)\sin x \cr
& \Rightarrow \,\,\sin \frac{x}{2}\left\{ {\sin \frac{x}{2} - \left( {\sqrt 2 - 1} \right)\cos \frac{x}{2}} \right\} = 0 \cr
& \therefore \,\,\sin \frac{x}{2} = 0\,\,\,\,{\text{or, }}\tan \frac{x}{2} = \sqrt 2 - 1 = \tan \frac{{{{45}^ \circ }}}{2} \cr
& \Rightarrow \,\,\frac{x}{2} = n\pi \,\,\,{\text{or, }}\frac{x}{2} = n\pi + \frac{\pi }{8} \cr
& \therefore \,\,x = 2n\pi ,2n\pi + \frac{\pi }{4}. \cr} $$