Trigonometric Equations Question 22
Question: If $ \tan \theta +\tan 2\theta +\sqrt{3}\tan \theta \tan 2\theta =\sqrt{3}, $ then
[UPSEAT 2001]
Options:
A) $ \theta =(6n+1)\pi /18,,\forall n\in I $
B) $ \theta =(6n+1)\pi /9,,\forall n\in I $
C) $ \theta =(3n+1)\pi /9,,\forall n\in I $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
- Given relation is $ \tan \theta +\tan 2\theta +\sqrt{3}\tan \theta \tan 2\theta =\sqrt{3} $
Þ $ ,\tan \theta +\tan 2\theta =\sqrt{3}(1-\tan \theta \tan 2\theta ) $
Þ $ \frac{\tan \theta +\tan 2\theta }{1-\tan \theta \tan 2\theta } $ = $ \sqrt{3} $
Þ $ \tan 3\theta =\tan (\pi /3) $
Þ $ 3\theta =n\pi +\frac{\pi }{3} $
Þ $ \theta =(3n+1)\frac{\pi }{9} $ .
BETA
