Trigonometric Equations Question 66
Question: The value of $ \theta $ in between $ 0^{o} $ and $ 360^{o} $ and satisfying the equation $ \tan \theta +\frac{1}{\sqrt{3}}=0 $ is equal to
[Pb. CET 2002]
Options:
A) $ \theta =150^{o} $ and $ 300^{o} $
B) $ \theta =120^{o} $ and $ 300^{o} $
C) $ \theta =60^{o} $ and $ 240^{o} $
D) $ \theta =150^{o} $ and $ 330^{o} $
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Answer:
Correct Answer: D
Solution:
- We have, $ \tan \theta +\frac{1}{\sqrt{3}}=0 $ or $ \tan \theta =-\frac{1}{\sqrt{3}} $ $ \because $ $ \theta $ lies in between $ 0{}^\circ $ and $ 360{}^\circ $
$ \therefore $ $ \theta =150{}^\circ $ and $ 330{}^\circ $ .
BETA
