question_answer

If\[\alpha ,\beta \in \left( 0,\frac{\pi }{2} \right),\sin \alpha =\frac{4}{5}\]\[\cos (\alpha +\beta )=-\frac{12}{13},\]then\[\sin \beta \]is equal to

A)  \[\frac{63}{65}\]                                             

B)  \[\frac{61}{65}\]

C)  \[\frac{3}{5}\]                  

D)         \[\frac{5}{13}\]

E)  \[\frac{8}{65}\]

Correct Answer: A

Solution :

Given that, \[\alpha ,\beta \in \left( 0,\frac{\pi }{2} \right)\]                 \[\sin \alpha =\frac{4}{5}\] \[\Rightarrow \]               \[\cos \alpha =\frac{3}{5}\] Now, \[\cos (\alpha +\beta )=-\frac{12}{13}\] \[\Rightarrow \]               \[\sin (\alpha +\beta )=-\frac{5}{13}\]    and \[cos\alpha \text{ }cos\beta -sin\alpha \text{ }sin\beta =-\frac{12}{13}\] \[\Rightarrow \] \[\frac{3}{5}\cos \beta -\frac{4}{5}\sin \beta =-\frac{12}{13}\]            ...(i) Now,      \[\sin (\alpha +\beta )=\frac{5}{13}\] \[\Rightarrow \]               \[\sin \alpha \sin \beta +\cos \alpha \sin \beta =\frac{5}{13}\] \[\Rightarrow \]               \[\frac{4}{5}\cos \beta +\frac{3}{5}sin\beta =\frac{5}{13}\]            ...(ii) Solving Eqs. (i) and (ii) simultaneously, we get                 \[\sin \beta =\frac{63}{65}\]