A) \[2n\pi \pm \frac{\pi }{4}+\frac{\pi }{12}\]
B) \[n\pi +{{(-1)}^{n}}\frac{\pi }{2}\]
C) \[2n\pi \pm \frac{\pi }{4}-\frac{\pi }{12}\]
D) None
Correct Answer: A
Solution :
Let \[\sqrt{3}+1=r\,\,\cos \alpha ,\]and \[\sqrt{3}-1=r\,\,sin\alpha \] \[\therefore \,\,{{r}^{2}}={{\left( \sqrt{3}+1 \right)}^{2}}+{{\left( \sqrt{3}-1 \right)}^{2}}=8\] i.e. \[\alpha =\pi /12\] From the equation, \[r\cos (\theta -\alpha )=2\] \[\Rightarrow \,\,\cos (\theta -\pi /12)=1/\sqrt{2}=\cos (\pi /4)\] \[\therefore \,\,\,\theta =2n\pi \pm \pi /4+\pi /12\]
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