Trigonometric Equations Question 71
Question: The most general value of $ \theta $ which will satisfy both the equations $ \sin \theta =-\frac{1}{2} $ and $ \tan \theta =\frac{1}{\sqrt{3}} $ is
[MNR 1980; MP PET 1989; DCE 1995]
Options:
A) $ n\pi +{{(-1)}^{n}}\frac{\pi }{6} $
B) $ n\pi +\frac{\pi }{6} $
C) $ 2n\pi \pm \frac{\pi }{6} $
D) None of these
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Answer:
Correct Answer: D
Solution:
- $ \sin \theta =-\frac{1}{2}=\sin ( -\frac{\pi }{6} )=\sin ( \pi +\frac{\pi }{6} ) $ $ \tan \theta =\frac{1}{\sqrt{3}}=\tan ( \frac{\pi }{6} )=\tan ( \pi +\frac{\pi }{6} )\Rightarrow \theta =( \pi +\frac{\pi }{6} ) $ Hence general value of $ \theta $ is $ 2n\pi +\frac{7\pi }{6} $ .
BETA
