Trigonometric Equations Question 386
Question: General value of $ \theta $ satisfying the equation $ {{\tan }^{2}}\theta +\sec 2\theta -=1 $ is
[IIT 1996]
Options:
A) $ m\pi ,n\pi +\frac{\pi }{3} $
B) $ m\pi ,n\pi \pm \frac{\pi }{3} $
C) $ m\pi ,n\pi \pm \frac{\pi }{6} $
D) None of these (Where m and n are integers)
Show Answer
Answer:
Correct Answer: B
Solution:
- Using $ \sec 2\theta =\frac{1}{\cos 2\theta }=\frac{1+{{\tan }^{2}}\theta }{1-{{\tan }^{2}}\theta } $ , we can write the given equation as $ {{\tan }^{2}}\theta +\frac{1+{{\tan }^{2}}\theta }{1-{{\tan }^{2}}\theta }=1 $ .
$ \Rightarrow $ $ {{\tan }^{2}}\theta (1-{{\tan }^{2}}\theta )+1+{{\tan }^{2}}\theta =1-{{\tan }^{2}}\theta $
$ \Rightarrow $ $ 3{{\tan }^{2}}\theta -{{\tan }^{4}}\theta =0\Rightarrow {{\tan }^{2}}\theta (3-{{\tan }^{2}}\theta )=0 $
$ \Rightarrow $ $ \tan \theta =0 $ or $ \tan \theta =\pm \sqrt{3} $ Now $ \tan \theta =0\Rightarrow \theta =m\pi $ , where m is an integer and tan $ \theta =\pm \sqrt{3}=\tan (\pm \pi /3) \Rightarrow \theta =n\pi \pm \frac{\pi }{3} $ , where n is an integer. Thus $ \theta =m\pi ,,n\pi \pm \frac{\pi }{3} $ , where m and n are integers.
BETA
