A) \[2n\pi \]
B) \[n\pi +{{(-1)}^{n}}\frac{\pi }{4}-\frac{\pi }{4}\]
C) \[2n\pi +\frac{\pi }{2}\]
D) None of these
Correct Answer: B
Solution :
\[\sin \theta +\cos \theta =1\Rightarrow \frac{1}{\sqrt{2}}\sin \theta +\frac{1}{\sqrt{2}}\cos \theta =\frac{1}{\sqrt{2}}\] Dividing by\[\sqrt{{{1}^{2}}+{{1}^{2}}}=\sqrt{2}\], we get \[\sin \left( \theta +\frac{\pi }{4} \right)=\frac{1}{\sqrt{2}}=\sin \frac{\pi }{4}\] \[\Rightarrow \] \[\theta +\frac{\pi }{4}=n\pi +{{(-1)}^{n}}\frac{\pi }{4}\Rightarrow \theta =n\pi +{{(-1)}^{n}}\frac{\pi }{4}-\frac{\pi }{4}\].
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