A) \[\frac{\sin \phi }{\sin \theta }\]
B) \[\frac{\sin \theta }{\sin \phi }\]
C) \[\frac{\sin \phi }{1-\cos \theta }\]
D) \[\frac{\sin \theta }{1-\cos \phi }\]
Correct Answer: B
Solution :
[b]: We have \[\tan \theta =\frac{x\sin \phi }{1-x\cos \phi }\] \[\Rightarrow \]\[x\sin \phi =\tan \phi -x\cos \phi \tan \theta \] \[\Rightarrow \]\[x=\frac{\tan \theta }{\sin \phi +\cos \phi \tan \theta }\] \[=\frac{\sin \theta }{\cos \theta \sin \phi +\cos \phi \sin \theta }=\frac{\sin \theta }{\sin (\theta +\phi )}\] Similarly,\[y=\frac{\sin \phi }{\sin (\theta +\phi )};\therefore \frac{x}{y}=\frac{\sin \theta }{\sin \phi }\].
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