A) \[2\,{{\sin }^{2}}A\]
B) \[\cos \,A\]
C) \[\sin \,2A\]
D) None of these
Correct Answer: D
Solution :
\[\frac{\cos \,2A-1}{\sin \,A}+\frac{\sin \,2A}{\cos \,2A+1}\times \cos \,A\] \[=\frac{1+{{\sin }^{2}}\,A-1}{\sin \,A}+\frac{\sin \,2A}{2\,{{\cos }^{2}}A-1+1}\times \cos \,A\] \[=\frac{2{{\sin }^{2}}A}{\sin \,A}+\frac{2\,\sin \,A\cdot \cos \,A}{2\,{{\cos }^{2}}A}\times \cos \,A\] \[=2\,\sin A+\sin \,A=3\,\sin \,A\]
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