Trigonometric Identities Question 373
Question: If $ \cos (\alpha +\beta )=\frac{4}{5},\sin (\alpha -\beta )=\frac{5}{13} $ and $ \alpha ,\beta $ lie between 0 and $ \frac{\pi }{4}, $ then $ \tan 2\alpha = $
[IIT 1979; EAMCET 2002]
Options:
A) $ \frac{16}{63} $
B) $ \frac{56}{33} $
C) $ \frac{28}{33} $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
We have $ \cos ,(\alpha +\beta )=\frac{4}{5} $ and $ \sin ,(\alpha -\beta )=\frac{5}{13} $
$ \Rightarrow \sin ,(\alpha +\beta )=\frac{3}{5} $ and $ \cos ,(\alpha -\beta )=\frac{12}{13} $
$ \Rightarrow 2\alpha ={{\sin }^{-1}}\frac{3}{5}+{{\sin }^{-1}}\frac{5}{13} $ $ ={{\sin }^{-1}}[ \frac{3}{5}\sqrt{1-\frac{25}{169}}+\frac{5}{13}\sqrt{1-\frac{9}{25}} ] $
$ \Rightarrow 2\alpha ={{\sin }^{-1}},( \frac{56}{65} ),\Rightarrow ,\sin ,2\alpha =\frac{56}{65} $ Now, $ \tan ,2\alpha =\frac{\sin ,2\alpha }{\cos ,2\alpha }=\frac{56/65}{33/65}=\frac{56}{33} $ .
BETA
