A) \[\frac{25}{16}\]
B) \[\frac{56}{33}\]
C) \[\frac{19}{12}\]
D) \[\frac{20}{7}\]
Correct Answer: B
Solution :
\[\tan 2\alpha =\tan \left[ (\alpha +\beta )+(\alpha -\beta ) \right]=\frac{\tan (\alpha +\beta )+\tan (\alpha -\beta )}{1-\tan (\alpha +\beta )\tan (\alpha -\beta )}\] as\[\cos (\alpha +\beta )=4/5,\sin (\alpha -\beta )=5/13\] \[\tan 2\alpha =\frac{\frac{3}{4}+\frac{5}{12}}{1-\frac{3}{4}.\frac{5}{12}}=\frac{\frac{9+5}{12}}{\frac{16-5}{16}}=\frac{56}{33}\]
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