A) \[\frac{63}{16}\]
B) \[\frac{61}{16}\]
C) \[\frac{65}{16}\]
D) \[\frac{32}{9}\]
Correct Answer: A
Solution :
\[0<\alpha <\frac{\pi }{4}\] |
\[0<\beta <\frac{\pi }{4}\] \[\Rightarrow \] \[0<\alpha <\beta <\frac{\pi }{2}\] |
And \[-\frac{\pi }{4}<\alpha -\beta <\frac{\pi }{4}\] |
Now \[\sin (\alpha -\beta )=\frac{5}{13}\Rightarrow \cos (\alpha -\beta )=\frac{12}{13}\] And \[\cos (\alpha +\beta )=\frac{3}{5}\] \[\Rightarrow \] \[sin(\alpha +\beta )=\frac{4}{5}\] |
Now\[\tan 2\alpha =\tan [(\alpha +\beta )+(\alpha -\beta )]\] |
\[=\frac{\tan (\alpha +\beta )+tan(\alpha -\beta )}{1-\tan (\alpha +\beta )tan(\alpha -\beta )}\] |
\[=\frac{\frac{4}{3}+\frac{5}{12}}{1-\frac{4}{3}\times \frac{5}{12}}=\frac{63}{16}.\] |
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