A) \[\frac{2bd}{{{b}^{2}}+{{d}^{2}}}\]
B) \[\frac{{{a}^{2}}+{{c}^{2}}}{2ac}\]
C) \[\frac{{{b}^{2}}+{{d}^{2}}}{2bd}\]
D) \[\frac{2ac}{{{a}^{2}}+{{c}^{2}}}\]
Correct Answer: D
Solution :
[d]: According to the given condition, \[\Rightarrow \]\[2\sin \frac{\alpha +\beta }{2}\cos \frac{\alpha -\beta }{2}=-a\] and\[2\cos \frac{\alpha +\beta }{2}\cos \frac{\alpha -\beta }{2}=-c\Rightarrow \tan \frac{\alpha +\beta }{2}=\frac{a}{c}\] Now,\[\sin (\alpha +\beta )=\frac{2\tan \left( \frac{\alpha +\beta }{2} \right)}{1+{{\tan }^{2}}\left( \frac{\alpha +\beta }{2} \right)}=\frac{2ac}{{{a}^{2}}+{{c}^{2}}}\]You need to login to perform this action.
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