Trigonometric Equations Question 337
Question: If $ \sqrt{2}\sec \theta +\tan \theta =1, $ then the general value $ \theta $ is
[MP PET 1989]
Options:
A) $ n\pi +\frac{3\pi }{4} $
B) $ 2n\pi +\frac{\pi }{4} $
C) $ 2n\pi -\frac{\pi }{4} $
D) $ 2n\pi \pm \frac{\pi }{4} $
Show Answer
Answer:
Correct Answer: C
Solution:
- $ \sqrt{2}\sec \theta +\tan \theta =1\Rightarrow \frac{\sqrt{2}}{\cos \theta }+\frac{\sin \theta }{\cos \theta }=1 $
$ \Rightarrow $ $ \sin \theta -\cos \theta =-\sqrt{2} $ Dividing by $ \sqrt{2} $ on both sides, we get $ \frac{1}{\sqrt{2}}\sin \theta -\frac{1}{\sqrt{2}}\cos \theta =-1 $
$ \Rightarrow $ $ \frac{1}{\sqrt{2}}\cos \theta -\frac{1}{\sqrt{2}}\sin \theta =1\Rightarrow \cos ,( \theta +\frac{\pi }{4} )=\cos (0) $
$ \Rightarrow $ $ \theta +\frac{\pi }{4}=2n\pi \pm 0\Rightarrow \theta =2n\pi -\frac{\pi }{4} $ .
BETA
