A) \[m\pi ,n\pi +\frac{\pi }{3}\]
B) \[m\pi ,n\pi \pm \frac{\pi }{3}\]
C) \[m\pi ,n\pi \pm \frac{\pi }{6}\]
D) None of these (Where m and n are integers)
Correct Answer: B
Solution :
Using\[\sec 2\theta =\frac{1}{\cos 2\theta }=\frac{1+{{\tan }^{2}}\theta }{1-{{\tan }^{2}}\theta }\], we can write the given equation as\[{{\tan }^{2}}\theta +\frac{1+{{\tan }^{2}}\theta }{1-{{\tan }^{2}}\theta }=1\]. \[\Rightarrow \] \[{{\tan }^{2}}\theta (1-{{\tan }^{2}}\theta )+1+{{\tan }^{2}}\theta =1-{{\tan }^{2}}\theta \] \[\Rightarrow \] \[3{{\tan }^{2}}\theta -{{\tan }^{4}}\theta =0\Rightarrow {{\tan }^{2}}\theta (3-{{\tan }^{2}}\theta )=0\] \[\Rightarrow \] \[\tan \theta =0\] or \[\tan \theta =\pm \sqrt{3}\] Now\[\tan \theta =0\Rightarrow \theta =m\pi \], where m is an integer and tan\[\theta =\pm \sqrt{3}=\tan (\pm \pi /3) \Rightarrow \theta =n\pi \pm \frac{\pi }{3}\], where n is an integer. Thus \[\theta =m\pi ,\,n\pi \pm \frac{\pi }{3}\], where m and n are integers.
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