A) \[n\pi +\frac{3\pi }{4}\]
B) \[2n\pi +\frac{\pi }{4}\]
C) \[2n\pi -\frac{\pi }{4}\]
D) \[2n\pi \pm \frac{\pi }{4}\]
Correct Answer: C
Solution :
\[\sqrt{2}\sec \theta +\tan \theta =1\Rightarrow \frac{\sqrt{2}}{\cos \theta }+\frac{\sin \theta }{\cos \theta }=1\] \[\Rightarrow \] \[\sin \theta -\cos \theta =-\sqrt{2}\] Dividing by \[\sqrt{2}\] on both sides, we get \[\frac{1}{\sqrt{2}}\sin \theta -\frac{1}{\sqrt{2}}\cos \theta =-1\] \[\Rightarrow \] \[\frac{1}{\sqrt{2}}\cos \theta -\frac{1}{\sqrt{2}}\sin \theta =1\Rightarrow \cos \,\left( \theta +\frac{\pi }{4} \right)=\cos (0)\] \[\Rightarrow \] \[\theta +\frac{\pi }{4}=2n\pi \pm 0\Rightarrow \theta =2n\pi -\frac{\pi }{4}\].
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