A) 0
B) 1
C) \[\sin \,y\]
D) \[\cos \,y\]
Correct Answer: D
Solution :
The expression can be written as \[\frac{1+\cos y-{{\sin }^{2}}y}{1+\cos y}+\frac{(1-{{\cos }^{2}}y)-{{\sin }^{2}}y}{\sin y\,(1-\cos y)}\] \[=\frac{\cos y\,(1+\cos y)}{1+\cos y}+0=\cos y.\]
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