Trigonometric Equations Question 368
Question: If $ \tan \theta +\tan 2\theta +\tan 3\theta =\tan \theta \tan 2\theta \tan 3\theta $ , then the general value of $ \theta $ is
Options:
A) $ n\pi $
B) $ \frac{n\pi }{6} $
C) $ n\pi -\frac{\pi }{4}\pm \alpha $
D) $ \frac{n\pi }{2} $
Show Answer
Answer:
Correct Answer: B
Solution:
- $ \tan \theta +\tan 2\theta +\tan 3\theta =\tan \theta \tan 2\theta \tan 3\theta $ $ \tan 6\theta =\frac{\tan \theta +\tan 2\theta +\tan 3\theta -\tan \theta \tan 2\theta \tan 3\theta }{1-\sum \tan \theta \tan 2\theta } $ = 0, (from the given condition)
$ \Rightarrow $ $ 6\theta =n\pi \Rightarrow \theta =n\pi /6 $ . Trick: In such type of problems, the general value of $ \theta $ is given by $ \frac{n\pi }{\text{sum of number of }\theta } $ . So the general value of $ \theta $ is $ \frac{n\pi }{1+2+3}=\frac{n\pi }{6} $ .
BETA
