A) \[2n\pi \pm \frac{\pi }{4}+\frac{\pi }{12}\]
B) \[n\pi +{{(-1)}^{n}}\frac{\pi }{4}+\frac{\pi }{12}\]
C) \[2n\pi \pm \frac{\pi }{4}-\frac{\pi }{12}\]
D) \[n\pi +{{(-1)}^{n}}\frac{\pi }{4}-\frac{\pi }{12}\]
Correct Answer: A
Solution :
Let\[\sqrt{3}+1=r\cos \alpha \] and\[\sqrt{3}-1=r\sin \alpha \] \[\therefore \]\[{{r}^{2}}{{\cos }^{2}}\alpha +{{r}^{2}}{{\sin }^{2}}\alpha ={{(\sqrt{3}+1)}^{2}}+{{(\sqrt{3}-1)}^{2}}\] \[\Rightarrow \] \[{{r}^{2}}=3+1+2\sqrt{3}+3+1-2\sqrt{3}\] \[\Rightarrow \] \[{{r}^{2}}=8\] \[\Rightarrow \] \[r=2\sqrt{2}\] and \[\tan \alpha =\frac{r\sin \alpha }{r\cos \alpha }=\frac{\sqrt{3}-1}{\sqrt{3}+1}\] \[=\frac{1-1/\sqrt{3}}{1+1/\sqrt{3}}\] \[=\frac{\tan \frac{\pi }{4}-\tan \frac{\pi }{6}}{1+\tan \frac{\pi }{4}.\tan \frac{\pi }{6}}\] \[=\tan \left( \frac{\pi }{4}-\frac{\pi }{6} \right)\] \[\Rightarrow \] \[\tan \alpha =\tan (\pi /12)\] \[\Rightarrow \] \[\alpha =\frac{\pi }{12}\] Given, equation is \[(\sqrt{3}-1)\sin \theta +(\sqrt{3}+1)\cos \theta =2\] \[\Rightarrow \] \[r\sin \alpha \sin \theta +r\cos \alpha \cos \theta =2\] \[\Rightarrow \] \[2\sqrt{2}\cos (\theta -\alpha )=2\] \[\Rightarrow \] \[\cos \left( \theta -\frac{\pi }{12} \right)=\frac{2}{2\sqrt{2}}=\cos \left( \frac{\pi }{4} \right)\] \[\Rightarrow \] \[\theta -\frac{\pi }{12}=2n\pi \pm \frac{\pi }{4}\] \[\Rightarrow \] \[\theta =2n\pi \pm \frac{\pi }{4}+\frac{\pi }{12}\]
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