A) \[\frac{21}{16}\]
B) \[\frac{63}{52}\]
C) \[\frac{33}{52}\]
D) \[\frac{63}{16}\]
Correct Answer: D
Solution :
\[0<\alpha +\beta =\frac{\pi }{2}\]and\[\frac{-\pi }{4}<\alpha -\beta <\frac{\pi }{4}\] if\[\cos (\alpha +\beta )=\frac{3}{5}\]then\[tan(\alpha +\beta )=\frac{4}{3}\] and if \[sin(\alpha -\beta )=\frac{5}{13}\]then\[tan(\alpha -\beta )=\frac{5}{12}\] (since \[\alpha -\beta \] here lies in the first quadrant) Now \[\tan (2\alpha )=tan\{(\alpha +\beta )+(\alpha -\beta )\}\] \[=\frac{\tan (\alpha +\beta )+\tan (\alpha -\beta )}{1-\tan (\alpha +\beta ).\tan (\alpha -\beta )}=\frac{\frac{4}{3}+\frac{5}{12}}{1-\frac{4}{3}.\frac{5}{12}}=\frac{63}{16}\]
You need to login to perform this action.
You will be redirected in
3 sec
