A) \[\frac{1+\tan \alpha }{1-\tan \alpha }\]
B) \[\frac{1+\sin \alpha }{1-\cos \alpha }\]
C) \[\frac{1+\sin \alpha }{1-\sin \alpha }\]
D) \[\frac{1+\cos \alpha }{1-\cos \alpha }\]
Correct Answer: A
Solution :
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