A) 0
B) \[-1\]
C) 1
D) 2
Correct Answer: A
Solution :
[a] \[\cos 2\phi +{{\sin }^{2}}\phi =\frac{1-{{\tan }^{2}}\phi }{1+{{\tan }^{2}}\phi }+{{\sin }^{2}}\phi \] \[=\frac{1-2{{\tan }^{2}}\phi -1}{1+2{{\tan }^{2}}\phi +1}+{{\sin }^{2}}\phi =\frac{-2{{\tan }^{2}}\phi }{2(1+{{\tan }^{2}}\phi )}+{{\sin }^{2}}\phi \] \[=\frac{\frac{{{\sin }^{2}}\phi }{{{\cos }^{2}}\phi }}{1+\frac{{{\sin }^{2}}\phi }{{{\cos }^{2}}\phi }}+{{\sin }^{2}}\phi =\frac{-{{\sin }^{2}}\phi }{{{\cos }^{2}}\phi +{{\sin }^{2}}\phi }+{{\sin }^{2}}\phi \] \[[\because {{\cos }^{2}}\phi +{{\sin }^{2}}\phi =1]\] \[=-si{{n}^{2}}\phi +si{{n}^{2}}\phi \] = 0
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