Trigonometric Identities Question 46
Question: If $ \tan \alpha =\frac{1}{7} $ and $ \sin \beta =\frac{1}{\sqrt{10}}( 0<\alpha ,,\beta <\frac{\pi }{2} ) $ , then $ 2\beta $ is equal to
Options:
A) $ \frac{\pi }{4}-\alpha $
B) $ \frac{3\pi }{4}-\alpha $
C) $ \frac{\pi }{8}-\frac{\alpha }{2} $
D) $ \frac{3\pi }{8}-\frac{\alpha }{2} $
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Answer:
Correct Answer: A
Solution:
Since $ \sin \beta =\frac{1}{\sqrt{10}}\Rightarrow \tan \beta =\frac{1}{3} $
Þ $ \tan 2\beta =\frac{2\tan \beta }{1-{{\tan }^{2}}\beta }=\frac{3}{4} $
$ \therefore \tan (\alpha +2\beta )=\frac{\frac{1}{7}+\frac{3}{4}}{1-\frac{1}{7}.\frac{3}{4}}=\frac{25}{25}=1 $ Now, $ 0<\beta <\frac{\pi }{2} $ and $ \tan 2\beta =\frac{3}{4}>0 $ both
Þ $ 0<2\beta <\frac{\pi }{2} $ . Again, $ 0<\alpha <\frac{\pi }{2} $ and $ 0<2\beta <\frac{\pi }{2} $ both
Þ $ 0<\alpha +2\beta <\pi $ Thus, $ 0<\alpha +2\beta <\pi $ and $ \tan (\alpha +2\beta )=1 $ both
Þ $ \alpha +2\beta =\frac{\pi }{4}\Rightarrow 2\beta =\frac{\pi }{4}-\alpha $ .
BETA
