A) \[\frac{\sin (\alpha +\beta )}{\sin (\alpha -\beta )}\]
B) \[\frac{\cos (\alpha -\beta )}{\cos (\alpha +\beta )}\]
C) \[\frac{\sin (\alpha -\beta )}{\sin (\alpha +\beta )}\]
D) \[\frac{\cos (\alpha +\beta )}{\cos (\alpha -\beta )}\]
Correct Answer: C
Solution :
\[{{\tan }^{2}}\frac{\theta }{2}=\frac{1-\cos \theta }{1+\cos \theta }=\frac{\tan \alpha -\tan \beta }{\tan \alpha +\tan \beta }=\frac{\sin (\alpha -\beta )}{\sin (\alpha +\beta )}\].
You need to login to perform this action.
You will be redirected in
3 sec
