A) \[\frac{\sin \varphi }{\sin \theta }\]
B) \[\frac{\sin \theta }{\sin \varphi }\]
C) \[\frac{\sin \varphi }{1-\cos \theta }\]
D) \[\frac{\sin \theta }{1-\cos \varphi }\]
Correct Answer: B
Solution :
We have \[\tan \theta =\frac{x\,\sin \,\varphi }{1-x\,\cos \,\,\varphi }\] \[\Rightarrow \,\,\frac{1}{x}\tan \theta -\tan \theta \,\,\cos \varphi =\sin \,\varphi \] \[\Rightarrow \,\,\frac{1}{x}=\frac{\sin \,\varphi +\cos \,\,\varphi \,\tan \,\theta }{\tan \,\theta }\] and \[\tan \,\varphi =\frac{y\,\sin \,\theta }{1-y\,\cos \,\theta }\] \[\Rightarrow \tan \,\varphi \,=\frac{\sin \,\theta }{\frac{1}{y}-\cos \,\theta }\]\[\Rightarrow \,\,\frac{1}{y}\tan \,\varphi -\tan \,\varphi \cos \theta =\sin \theta \] \[\Rightarrow \,\,\frac{1}{y}\tan \,\varphi \,=\sin \,\theta +\tan \,\varphi \,\cos \theta \] \[\therefore \,\,\,\,\frac{1}{y}=\frac{\sin \,\theta +\tan \,\varphi \,\cos \theta }{\tan \,\varphi }\] Now \[\frac{x}{y}=\left[ \frac{\tan \,\theta }{\sin \,\varphi +\cos \,\varphi \,\tan \,\theta } \right]\times \left[ \frac{\sin \,\theta +\tan \,\varphi \,\cos \,\theta }{\tan \,\varphi } \right]\] \[=\frac{\tan \,\theta }{\tan \,\varphi }\,\left[ \frac{\sin \,\theta +\cos \,\theta \frac{\sin \varphi }{\cos \varphi }}{\sin \varphi +\cos \varphi \frac{\sin \theta }{\cos \theta }} \right]=\frac{\tan \theta \,\,\cos \theta }{\tan \,\varphi \,\cos \varphi }=\frac{\sin \theta }{\sin \varphi }\] Aliter: \[x\,\sin \,\varphi =\tan \,\theta -x\,\cos \,\varphi \,\tan \,\theta \] \[\Rightarrow \,x=\frac{\tan \,\theta }{\sin \,\varphi +\cos \,\varphi \,\tan \,\theta }\] \[=\frac{\sin \,\theta }{\cos \,\theta \sin \,\varphi +\cos \,\varphi \,\sin \,\theta }=\frac{\sin \,\theta }{\sin \,(\theta +\varphi )}\] Similarly, \[y=\frac{\sin \,\varphi }{\sin \,(\theta +\varphi )}\]; \[\therefore \,\,\frac{x}{y}=\frac{\sin \theta }{\sin \varphi }.\]
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