A) \[\frac{n\pi }{3}-\frac{\pi }{12}\]
B) \[n\pi +\frac{7\pi }{12}\]
C) \[\frac{n\pi }{3}+\frac{7\pi }{36}\]
D) \[n\pi +\frac{\pi }{12}\]
Correct Answer: C
Solution :
| Given \[\frac{\tan 3\theta -1}{\tan 3\theta +1}=\sqrt{3}\] |
| \[\Rightarrow \,\,\,\sqrt{3}(\tan 3\theta +1)=\tan 3\theta -1\] |
| \[\Rightarrow \,\,\,\sqrt{3}\,\tan 3\theta +\sqrt{3}=\tan 3\theta -1\] |
| \[\Rightarrow \,\,\,\sqrt{3}\,\tan 3\theta -\tan 3\theta +1+\sqrt{3}=0\] |
| \[\Rightarrow \,\,\,\tan 3\theta (\sqrt{3}-1)+(1+\sqrt{3})=0\] |
| \[\Rightarrow \,\,\tan \theta (\sqrt{3}-1)=-(1+\sqrt{3})\] |
| \[\Rightarrow \,\,\tan 3\theta =\frac{-(\sqrt{3}+1)}{(\sqrt{3}-1)}=\frac{-(1+\sqrt{3})}{-(1-\sqrt{3})}=\frac{1+\sqrt{3}}{1-\sqrt{3}}\] |
| \[\Rightarrow \,\,\tan 3\theta =\tan 105{}^\circ =\tan \frac{7\pi }{12}\] |
| [Note: \[\tan \theta =\tan \alpha \Rightarrow \theta =n\pi +\alpha \]] |
| \[\therefore \,\,3\theta =n\pi +\frac{7\pi }{12}\Rightarrow \theta =\frac{n\pi }{3}+\frac{7\pi }{36}\] |
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