Trigonometric Equations Question 369
Question: The solution of $ 3\tan (A-15^{o})=\tan (A+15^{o}) $ is
Options:
A) $ n\pi +\frac{\pi }{4} $
B) $ 2n\pi +\frac{\pi }{4} $
C) $ 2n\pi -\frac{\pi }{4} $
D) $ \frac{n\pi }{2}+{{(-1)}^{n}}\frac{\pi }{2} $
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Answer:
Correct Answer: A
Solution:
- $ \frac{3\sin (A-15^{o})}{\cos (A-15^{o})}=\frac{\sin (A+15^{o})}{\cos (A+15^{o})} $
Þ $ 3\sin (A-15^{o})\cos (A+15^{o}) $ $ =\cos (A-15^{o})\sin (A+15^{o}) $
$ \Rightarrow $ $ 2\sin (A-15^{o})\cos (A+15^{o})=\frac{1}{2} $
$ \Rightarrow $ $ \sin 2A-\sin 30^{o}=\frac{1}{2} $
$ \Rightarrow $ $ 2A=2n\pi +\frac{\pi }{2} $
$ \Rightarrow $ $ A=n\pi +\frac{\pi }{4} $ .
BETA
