A) \[n\pi +\frac{\pi }{8}\]
B) \[2n\pi ,\,\,2n\pi +\frac{\pi }{4}\]
C) \[2n\pi \]
D) None of these
Correct Answer: B
Solution :
Given, \[\sec x-1=(\sqrt{2}-1)\tan x\] \[\Rightarrow \] \[1-\cos x=(\sqrt{2}-1)\sin x\] \[\Rightarrow \] \[\sin \frac{x}{2}\left( \sin \frac{x}{2}-(\sqrt{2}-1)\cos \frac{x}{2} \right)=0\] \[\Rightarrow \]\[\sin \frac{x}{2}=0\] or \[\tan \frac{x}{2}=\sqrt{2}-1=\tan \frac{{{45}^{o}}}{2}\] \[\Rightarrow \] \[\frac{x}{2}=n\pi \]or\[\frac{x}{2}=n\pi +\frac{\pi }{8}\] \[\Rightarrow \] \[x=2n\pi ,\,\,2n\pi +\frac{\pi }{4}\]
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