A) \[\frac{\sin \theta }{1-\cos \theta }\]
B) \[\frac{\sin \theta }{1-\cos \phi }\]
C) \[\frac{\sin \phi }{\sin \theta }\]
D) \[\frac{\sin \theta }{\sin \phi }\]
Correct Answer: D
Solution :
| [d] We have,\[\tan \theta =\frac{a\sin \phi }{1-a\cos \phi }\] |
| \[\cot \theta =\frac{1}{a\sin \phi }-\cot \phi \] |
| \[\cot \theta +\cot \phi =\frac{1}{a\sin \phi }\] (1) |
| \[\tan \phi =\frac{b\sin \theta }{1-b\cos \theta }\] |
| \[\cot \theta =\frac{1}{b\sin \theta }-\cot \theta \] |
| \[\cot \phi +\cot \theta =\frac{1}{b\sin \theta }\] |
| From eqs. (1) and (2). we have |
| \[\frac{1}{a\sin \phi }=\frac{1}{b\sin \theta }\] |
| \[\frac{a}{b}=\frac{\sin \theta }{\sin \phi }\] |
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