A) \[n\pi \pm \frac{\pi }{6}\]
B) \[n\pi +\frac{\pi }{6}\]
C) \[2n\pi \pm \frac{\pi }{6}\]
D) None of these
Correct Answer: A
Solution :
\[\frac{1-{{\tan }^{2}}\theta }{{{\sec }^{2}}\theta }=\frac{1}{2}\Rightarrow {{\cos }^{2}}\theta -{{\sin }^{2}}\theta =\frac{1}{2}\] \[\Rightarrow \] \[\cos 2\theta =\frac{1}{2}=\cos \left( \frac{\pi }{3} \right)\] \[\Rightarrow \] \[2\theta =2n\pi \pm \frac{\pi }{3}\Rightarrow \theta =n\pi \pm \frac{\pi }{6}\].
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