Trigonometric Equations Question 363
Question: If $ \tan \theta -\sqrt{2}\sec \theta =\sqrt{3} $ , then the general value of $ \theta $ is
Options:
A) $ n\pi +{{(-1)}^{n}}\frac{\pi }{4}-\frac{\pi }{3} $
B) $ n\pi +{{(-1)}^{n}}\frac{\pi }{3}-\frac{\pi }{4} $
C) $ n\pi +{{(-1)}^{n}}\frac{\pi }{3}+\frac{\pi }{4} $
D) $ n\pi +{{(-1)}^{n}}\frac{\pi }{4}+\frac{\pi }{3} $
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Answer:
Correct Answer: B
Solution:
- $ \sin \theta +\cos \theta =\sqrt{2}\cos \alpha \Rightarrow \cos ( \theta -\frac{\pi }{4} )=\cos \alpha $
$ \Rightarrow $ $ \theta -\frac{\pi }{4}=2n\pi \pm \alpha \Rightarrow \theta =2n\pi +\frac{\pi }{4}\pm \alpha $ .
BETA
