Trigonometric Equations Question 64
Question: If $ \sin \theta =\sqrt{3}\cos \theta ,-\pi <\theta <0 $ , then $ \theta = $
[MP PET 1992]
Options:
A) $ -\frac{5\pi }{6} $
B) $ -\frac{4\pi }{6} $ $ $
C) $ \frac{4\pi }{6} $
D) $ \frac{5\pi }{6} $
Show Answer
Answer:
Correct Answer: B
Solution:
- $ \tan \theta =\sqrt{3}=\tan \frac{\pi }{3}\Rightarrow \theta =n\pi +\frac{\pi }{3} $ For $ -\pi <\theta <0 $ $ Putn=-1 $ , we get $ \theta =-\pi +\frac{\pi }{3}=\frac{-2\pi }{3}\text{or }\frac{-4\pi }{6} $ .
BETA
